Find the limit (n->inf for the following limit)

[dusty horizon]

1. Do not ping the Moderators, unless someone is breaking the rules.
2. Do not ping the Helper Moderators, unless there is a conflict between helpers.
3. Do not ping other members randomly for help.
4. Ask your question and show the work you've done so far. If you've posted a screenshot of a question, specify which part you need help with.
5. Wait patiently for a helper to come along.
6. If the Helper has answered your question, remember to thank them with the Mathematics Ranks bot and close the thread with:

+close
Feel free to nominate the person for helper of the week in #helper-nominations
If you're happy with the help you got here, and the server overall, you can contribute financially as well:

[lost badger]

This is what i've done so far

nvm now that I'm looking at it what I did is wrong

since a is just a constant value

[civic ember]

this doesn't make sense

there is no series here anywhere

this is a sequence

$$ x_n = \frac{1}{\ln n} \sum _{k=0}^n \frac{1}{a^k} $$

[echo tapir]

aL

[lost badger]

okay wait I think i misinterpreted the question

[civic ember]

the sum is geometric provided a >1

so maybe that helps you somehow

[fickle granite]

you can compute the sum explicitly

I don't see a reason not to do it here

[lost badger]

is a_n not a partial sum of an infinite series?

i'm trying to understand the whole thing

[civic ember]

there is no series here

[lost badger]

1+1/a+...+1/a^n is a sum though

[civic ember]

yes, a finite sum

[fickle granite]

a sum with n+1 terms which has another explicit form

you can turn it into a sum of 2 terms

[lost badger]

i'm confused here

[civic ember]

es una serie geometrica

[lost badger]

for n->inf is it not the same as rewriting it as sum of 1/a^(k-1)

[civic ember]

$$ \sum _{k=0}^\infty a_k \quad\textbf{is a series} $$

[echo tapir]

aL

[lost badger]

a_1 = 1/a^0 = 1
a_2 = 1/a^1 = 1/a
...

[civic ember]

$$ \frac{1}{\ln n} \sum _{k=0}^n a_k\quad\textbf{ is a sequence} $$

[echo tapir]

aL

[lost badger]

it was meant to be an n, not an inf above the sigma

idk why i wrote inf

[civic ember]

follow rion's suggestion

[lost badger]

i don't understand what he means

[civic ember]

what is 1+x+x^2 for example

$$ \frac{x^3-1}{x-1} = 1+x+x^2 $$

[echo tapir]

aL

[civic ember]

recognise this?

[lost badger]

oh is this a geometric series

[civic ember]

geometric sum

[lost badger]

sum yeah sorry

[fickle granite]

I believe you should know how to simplify a geometric sum

[civic ember]

write this in said form

think of x as 1/a

so you have 1+x+x^2+...+x^n

[lost badger]

can't find how to simplify this in my notes, i apologise

[civic ember]

well, you have to learn these things sooner or later

so you may as well learn them now

[fickle granite]

Please look for it, and learn it by heart

(well the proof is important too, but the formula is important enough to be remembered)

sorry that's a bad one

https://mathworld.wolfram.com/GeometricSeries.html

See from equation (1) to (6), which are applicable for any value of r except 1

[lost badger]

i'm confused on how to interpret this, is this then for r=1/a?

[civic ember]

correct

[lost badger]

i found the criteria in my notes and i'm confused then for this

[fickle granite]

It's a bit unrelated to our problem

[lost badger]

oh

[civic ember]

it's about convergence of the geometric series

[fickle granite]

It basically tells you what happens when n goes to infinity

but here we just want to simplify the sum

[civic ember]

it comes into play only for when we calculate the limit

[lost badger]

yeah i don't have anything directly related to the sum itself

[civic ember]

you can work it out by hand

[fickle granite]

Yeah, that's alright. I still kindly ask you to learn the identity I linked above, it is really important

[lost badger]

and another question, am i not supposed to input k=1 for the sum instead of k=0?

[civic ember]

doesn't really matter but k=0 right now

[fickle granite]

k starting from zero ensures the first term is 1

[lost badger]

right

[fickle granite]

(incidentally that's why I deleted the first link I posted, the sum started at k = 1)

[lost badger]

so then how is it meant to be rewritten? i'm confused here

[civic ember]

$$\frac{x^{n+1}-1}{x-1} $$

[echo tapir]

aL

[civic ember]

give it some thought, you'll see why

no doubt you have seen this in high school already when you calculated sum of geometric progression

[lost badger]

is a sum of geometric progression not a geometric series what

[civic ember]

series = infinite sum

not relevant right now

[lost badger]

and no i've never taken this

like ever

[civic ember]

of course you haven't

[civic ember]

regardless this here is the sum

replace x = 1/a and calculate the limit

[lost badger]

so the sum of (1/a)^k = (r^(n+1) -1) / (r-1) ?

[civic ember]

yes, for r = 1/a

[lost badger]

yeah i just couldn't be bothered to input 1/a for every r

is there a specific name for this identity

[civic ember]

geometric sum

we said this already

suma geometrica

[lost badger]

yeah i cant find anything of the like not even on google

like as for the identity

[civic ember]

yes. and others can

calculate the limit now

[lost badger]

i assume it's just this

because i don't understand

[civic ember]

you're not supposed to understand within the first 5 minutes of looking at it

you're supposed try and solve the practice problems and then you will understand

same way I could say "I don't understand" when I'm first learning spanish

I'm not supposed to! that's why we learn and practice

you have everything now to calculate limit of the sequence x_n

[lost badger]

so double checking

this is right no?

[civic ember]

and also the term 1/log n

[lost badger]

yes but this is for a_n only

[civic ember]

this is correct

[lost badger]

then lim n->inf ((1/a)^(n+1)-1) / ((1/a) - 1 ) · log(n)

[civic ember]

1/log n

[lost badger]

like this

yeah i don't understand how to proceed otherwise

[civic ember]

almost there now

what is the limit?

consider two cases

a <= 1 and a > 1

[fickle granite]

Hang in there, you're close to the goal

[lost badger]

wouldn't it be different for a=1 and a<1 though

[civic ember]

you tell me

try a=1

what do you get?

[lost badger]

i would assume so because 1^(n+1) = 1^inf

[civic ember]

1^inf is not a thing

forget about it

don't ever write it

[lost badger]

while for a<1, a^(n+1) tends to 0 for n->inf

so you get 1/0, k/0 = inf

[civic ember]

nope

it's correct that for a <= 1 the sequence has infinite limit

but your reasoning is not correct

[civic ember]

this makes no sense at all

For a = 1 you have the following

[lost badger]

wait oops i mistook it for k/inf = 0

which IS correct

[civic ember]

$$ \lim x_n = \lim\frac{n+1}{\ln n} $$

[echo tapir]

aL

[lost badger]

i'm confused how did you simplify this

[civic ember]

i didn't simplify it

for a=1 you use this

[civic ember]

if you try plugging in a=1 there you just get 0/0 which is nonsense

$$\lim x_n = \lim\frac{n+1}{\ln n} \geqslant \lim \frac{n}{\ln n} = \infty $$

[echo tapir]

aL

[civic ember]

if a < 1, by the same reasoning you conclude the limit is inf

[lost badger]

i just dont understand the way you developed it

is this not easier to solve by using stolz

yeah idk i can't reach an answer regardless

[civic ember]

using what?

elaborate

[lost badger]

stolz-cesaro theorem

[civic ember]

why

[lost badger]

since b_n = log(n), lim n->inf log(n) = inf

[civic ember]

maybe you could, but it's certainly not "easier"

[lost badger]

i mean i can't arrive at a solution with either method

[civic ember]

so you mean to tell me you are able to apply something like stolz cesaro and Not understand the basics of the problem?

just no

tell me why does stolz cesaro theorem apply in the first place?

[lost badger]

for {b_n} is strictly increasing (log(n) is strictly increasing) and lim n->inf {b_n} = inf you can apply it

[civic ember]

and a_n?

[lost badger]

since lim n->inf (a_n+1 - a_n / b_n+1 - b_n) = l (whether it be finite or infinite) = lim n->inf (a_n/b_n)

[civic ember]

and why does this limit exist?

[lost badger]

a_n, b_n only have to be sequences of real numbers

[civic ember]

and in this problem, what do you take as a_n?

[lost badger]

1+1/a+...+1/a^n

[civic ember]

so you are left with what

[lost badger]

unless i did it wrong i ended up with lim (n->inf) 1/ log(1+1/n) · a^(n+1)

[civic ember]

and you're saying this is easier

have you even proved stolz cesaro?

[lost badger]

we're not required to

[civic ember]

[lost badger]

well i honestly am lost in general

so

[civic ember]

what you are doing is hunting a bug with a tank

[civic ember]

and yet you are somehow magically able to make sense of stolz cesaro

no, I don't buy it

i call bullshit, I'm sorry

[lost badger]

i dunno what to tell you

let me explain thoroughly what i dont understand

[civic ember]

please do

[lost badger]

so from here

i dont understand how you go from this

[lost badger]

to this

[civic ember]

I told you, for a=1 I don't use this formula because it's 0/0 which is nonsense

for a=1 you simply have 1+1+...+1 / log n

[lost badger]

i mean the reason i avoided that is because our teacher has strictly prohibited us from using l'hopital for limits of sequences

[civic ember]

im not using lhopital

[lost badger]

yes i meant because

i arrived at 0/0

[lost badger]

okay, now i understand this

but an "infinite" sum of 1 is infinite?

[civic ember]

there is no infinite sum

i keep telling you

you have a sequence x_n and for each n the sum in the sequence is finite

[lost badger]

this is in the limit n->inf though

[civic ember]

still no infinite sum

[lost badger]

okay, regardless

so then

1+1+...+1 / log (n)

[civic ember]

yes

[lost badger]

in the limit you're left with inf in the denominator

then..?

[civic ember]

can you show without lhopital that this is true

$$ \lim \frac{n}{\ln n}=\infty $$

[echo tapir]

aL

[fickle granite]

This is a known, usual limit

just like how the limit of exp(n)/n is known

[civic ember]

where do you get these infs? :S

https://tenor.com/view/jim-carrey-frustrated-pulling-hair-no-frustrating-gif-14800061

[fickle granite]

you normally don't have to prove it again, but should you not know it, I also urge you to learn those usual limits (they are also as important as the geometric sum, and high school material)

[lost badger]

i assume i just have a terrible understanding of sequences in general but

log (n)

lim n->inf (log(n)) = inf

[civic ember]

that is true

log n is unbounded

[fickle granite]

That is correct, yes, but what happens if you divide it by n?

[lost badger]

if then you have that in the denominator how is it possible there's no inf in the denominator

[civic ember]

oh I see what's going on

you assume this right

$$ \lim \frac{a_n}{b_n} = \frac{\lim a_n}{\lim b_n}$$

[echo tapir]

aL

[lost badger]

yes

[civic ember]

that is true if the limits are finite

and lim b_n is nonzero obviously

[fickle granite]

or lim b_n (exists and) is nonzero, finite and lim a_n is infinite

[lost badger]

so then how am i meant to proceed is my question

[civic ember]

well I did tell you to look at two cases

let's say we figured out a=1

[lost badger]

yeah lim n->inf (1+...+1 / log(n))

[civic ember]

limit is inf

[lost badger]

i still don't understand exactly why is this limit = inf

[civic ember]

$$ \frac{n}{\ln n} $$

[echo tapir]

aL

[civic ember]

because this sequence is unbounded

[lost badger]

wait idk if we're on the same line

it's not ln in the bottom but log of base 10

[civic ember]

doesn't make a difference

result is the same

[lost badger]

i just realised by using l'hopitals rule lim n->inf for this is just 1/(1/n)

yeah okay proceeding

so lim = inf

[civic ember]

now let 0< a < 1

then 1/a > 1

[lost badger]

yes. then the numerator tends to infinite for lim n->inf

[civic ember]

$$ \frac{1}{\log n} \sum _{k=0}^n \frac{1}{a^k} \geqslant \frac{n+1}{\log n} \geqslant \frac{n}{\log n} $$

[echo tapir]

aL

[civic ember]

what do we conclude?

[lost badger]

i still don't understand why 1/log(n) · sum

[lost badger]

oh wait nvm

sorry

[civic ember]

that's how the sequence is defined

[lost badger]

yeah i was just not thinking

[civic ember]

so, what do we conclude for a < 1

[lost badger]

i just don't get why larger than for the first two

[civic ember]

1/a > 1, yes?

[lost badger]

yes, for 0<a<1

[civic ember]

so sum 1/a^k > 1 +1+1+...+1 , yes?

[lost badger]

for a=1 yes..?

but not for 0<a<1

[civic ember]

for a < 1

dont stop thinking please

[lost badger]

? i'm confused

[civic ember]

each term in the sum is greater than 1

because 1/a > 1

[lost badger]

oh wait

sorry

i read it wrong

🤦 i read a = not a >

i apologise

[civic ember]

do you believe this now?

[lost badger]

yes

[civic ember]

so what is the conclusion

lim x_n = ?

[lost badger]

inf

[civic ember]

correct

and now if a > 1, you have a geometric sum

and that is bounded for all n

hence the limit of x_n is 0

[lost badger]

how is the limit 0 i don't get it

[civic ember]

$$ \sum _{k=0}^n \frac{1}{a^k} $$

[echo tapir]

aL

[civic ember]

this sequence is bounded

now we can talk about series

the limit of this sequence is the sum of the geometric series which is finite, because a > 1

lim 1/log n =0

hence the limit of the product of the two is 0

[civic ember]

if I had 20 cents every time you said "you don't get it" after a few minutes of looking at something new I wouldn't need to go to work anymore

[civic ember]

.

you have to be able to solve this stuff on your own

and spoiler alert, this stuff is very basic, you will no doubt have more difficult problems to solve in your class

and don't test your teachers patience with saying something like "isn't it easier to apply stolz cesaro"

I promise you that won't end well

[lost badger]

i didn't mean to ever come off as rude whatsoever but not one single time has our teacher developed a problem like this

[civic ember]

and you also said you had been ill and not taken part in the classes

[lost badger]

yeah

[civic ember]

so maybe don't speak for your teacher if you weren't there

[lost badger]

🤷

[civic ember]

take responsibility, it's not your teacher's job to understand it for you

[lost badger]

i just don't understand certain stuff of this sort because i've never seen something of the like

[civic ember]

because this stuff is new to you

tell me why do you want to understand it in like the first few hours of seeing it?

or worse, the first 10 minutes?

[lost badger]

i don't really have the time to understand it within hours either

our first exams are coming up and all practise i've got is through this

[civic ember]

well, buckle up buddy, I told you there will be much harder exercises in the future

[lost badger]

i understand if i may have come off as impolite which was never my intention

i'm just very confused in general

[civic ember]

not impolite, but very impatient

[lost badger]

for this, so |r|<1 then S always converges to a finite sum?

[civic ember]

r=1/a, yes

[lost badger]

right, so this is why for a<1 it diverges?

i mean the sum

for any a<1, a=/=0

[civic ember]

yes, the geometric series converges if and only if |r|<1

[lost badger]

hope you don't mind me asking but it's something that confuses me since i started seeing series and any sort of (infinite) sums

okay nvm i don't know how to phrase it properly

[civic ember]

you can say in spanish

[lost badger]

like how is it possible the sum of r here converges if it's still an endless addition of 1+1/a+1/a^2+...+1/a^n?

is this something i'm just supposed to acknowledge? because i tried digging for proof and i couldn't understand it well

[civic ember]

that's a bad way to think about it, as an "endless sum"

that is not what a series is

$$ \sum _{k=0}^\infty a_k $$

[echo tapir]

aL

[civic ember]

this symbol doesn't mean it's an endless sum

it's just an informal intuition

that we write a_0 + a_1 + a_2 + ...

formally, the sum of the series is defined like this

$$\sum _{k=0}^\infty a_k = \lim _m \sum _{k=0}^m a_k $$

[echo tapir]

aL

[civic ember]

these are finite sums

for every m

this is called the sequence of partial sums

and if this sequence has a finite limit, then we say the series converges

[lost badger]

so it never really is an "infinite" sum

[civic ember]

formally no

it's a limit of a sequence of finite sums

and I am 105% certain that all of this is included in your lecture notes

[lost badger]

so sorry to disappoint but unfortunately no

[civic ember]

I don't believe that

so, here's what you should do going forward

[lost badger]

regardless it's better that i know this now than never

[civic ember]

learn the definitions

definitions and examples

without definitions, as you have already seen, you have nothing

and remember, just because you have the "right answer" doesn't mean your solution is correct

alright that's all for me

buenas noches y buena suerte

[lost badger]

thanks

[lost badger]

as a question related to this same exercise

through stolz-cesaro theorem I arrive at this limit

$\lim_{n \to \infty} \frac{1}{log(1+\frac{1}{n}) \cdot a^{n+1}}$

[echo tapir]

lucho

[lost badger]

would it be appropriate to simplify log(1+1/n) = 1/n since for log(1+x) = x for sufficiently small x?

so then the limit would end as

$\lim_{n \to \infty} \frac{n}{a^{n+1}}$

[echo tapir]

lucho

[lost badger]

after applying l'hopital, $\lim_{n \to \infty} \frac{1}{n \cdot a^{n}}$ if i'm not wrong?

[echo tapir]

lucho

[lost badger]

however for |a|<1 if you were to plug in n->inf this would be yet another 0 times inf scenario so not really useful

[fickle granite]

The justification here is not too rigorous

You could say that ln(1 + 1/n) = (1/n) × n ln(1 + 1/n)

And now you basically know that the right factor has a limit

Which allows you to deal with the rest

[lost badger]

it's not ln

or does this also apply to log base 10

[fickle granite]

You still get the point

Also log10(1 + 1/n) is not equivalent to 1/n

[lost badger]

oh then i misread it

what would be the way to simplify it then?

if you don't mind me asking, that is

or rewrite it in a way that doesn't make me end up with log_10 (1) = 0

[fickle granite]

$\frac{1}{n \ln(1 + 1/n)} \frac{n}{a^{n+1}}$

[echo tapir]

Rion

[fickle granite]

Also that is only assuming this is correct, which I will leave to your discretion

[lost badger]

sorry again about asking basically the exact same thing, but does it really matter whether it's ln(1+1/n) or log_10 (1+1/n) for here

[fickle granite]

n log10(1 + 1/n) does not converge to 1

[lost badger]

i'm confused about what's your point here, sorry

i mean yeah it's not the same i know

[fickle granite]

That your initial assumption is wrong, log10(1 + x) is not equivalent to x

[lost badger]

yeah i messed that up, that's my fault on my end

i mistook log10 for ln as it was solely written log

[fickle granite]

Also to be clear ln and log are synonymous to me

The natural logarithm

[lost badger]

yeah, that's where the issue came from

i'm interpreting log = log10 here

[fickle granite]

My bad

[lost badger]

no issue

[lost badger]

so then what could be a proper way to rewrite this?

since i only have an issue for a=1, 0<a<1

[fickle granite]

Ping me again in 30 minutes, I need to walk home

[lost badger]

okay, thank you

[lost badger]

@fickle granite

[fickle granite]

Assuming that it is in fact correct

[lost badger]

this is where i've arrived with s-c

[fickle granite]

$$\lim_{n \to \infty} \frac{1}{\ln\left(1 + \frac{1}{n}\right) a^{n+1}} = \lim_{n \to \infty} \frac{1}{n \ln\left( 1 + \frac{1}{n} \right)} \frac{n}{a^{n+1}}$$

[echo tapir]

Rion

[fickle granite]

This is true because both expressions are in fact the same thing

[lost badger]

yes, only issue i have with this expression is the log10 in the bottom

since 1/n will be 0 for n->inf. log10 (1) = 0

this is not a problem for 0<a<1 since a^n+1 will still converge to 0

[fickle granite]

yes but n log10(1+1/n) = n ln(1+1/n)/ln(10)

and what is the limit of n ln(1 + 1/n) ?

Hint: what is the definition of the deriative of ln at the point 1?

[lost badger]

should this not be just 0 still

[fickle granite]

no

[lost badger]

wait no

1

sorry

[fickle granite]

yes

so divide that by ln(10)

and you have that $\frac{1}{n \log_{10}\left(1 + \frac{1}{n} \right)} \to \ln(10)$

[echo tapir]

Rion

[lost badger]

i mean obviously this is still a finite number and no longer 0, so i no longer have the indeterminate form in the denominator

unless im just misinterpreting this

[fickle granite]

Which basically tells you that the convergence problem is limited to the other factor

[lost badger]

yes. only question i have here though is

since a^(n+1) is dependant on values of a, does a=1 not offer a problem here?

yes, i still have the issue with properly understanding the indeterminate forms

[fickle granite]

indeed, if a = 1, then you clearly have the other factor is just n

then, the limit would be the limit of n times ln(10), which you can easily conclude for

(though I don't know if you can apply Stolz Cesaro if the limit is infinite then)

[lost badger]

n·1/ln(10) here no?

[fickle granite]

the denominator tends to 1/ln(10)

so 1/denominator tends to ln(10)

[fickle granite]

this

[lost badger]

hold on i don't know if im misinterpreting this

let me rewrite it with texit

[fickle granite]

If you do, do me a favor and put your text in a verbatim block

[lost badger]

$n log10(1+\frac{1}{n}) = n \frac {ln(1+\frac{1}{n})}{ln(10)}$

oh sorry

[echo tapir]

lucho

[fickle granite]

See the problem is that Discord formats the subscripts

so if you copy paste them

you lose them

same for other special characters

unless the text is completely verbatim

$n \log_{10}(1+\frac{1}{n}) = n \frac {\ln(1+\frac{1}{n})}{\ln(10)}$

[echo tapir]

Rion

[lost badger]

alright

[fickle granite]

Yes, this is true

[lost badger]

wait now that i'm double checking this, i'm only able to deduce the numerator in the second expression = 1 because i'm taking the limit independently for the numerator and rewriting n as 1/n in the denominator and then using l'hopital

is there a more "simple" or "intuitive" way to deduce this?

or is this something i should just learn by heart?

like

[fickle granite]

n ln(1 + 1/n) tends to 1, I thought we established that

It's useful to know that

[lost badger]

$\lim_{n \to \infty} {n \cdot ln(1+\frac{1}{n})} = \lim_{n \to \infty} \frac {n}{n+1}

i think i wrote it right hold on

[echo tapir]

lucho
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[fickle granite]

It stems from the fact that:
$$\ln'(1) = \frac{1}{1}$$
Because $\ln'(x) = 1/x$, but also:
$$\ln'(1) = \lim_{x \to 0} \frac{\ln(1+x) - \ln(1)}{x}$$

[echo tapir]

Rion

[fickle granite]

you don't need l'Hospital for this

[lost badger]

ohhh

i didn't think about it this way

[lost badger]

this is just 1

[fickle granite]

Yes

[lost badger]

so then this is just 1/ln(10) correct

[fickle granite]

Therefore the limit of this guy is also known

yes

now you take the reciprocal of that

[lost badger]

okay, so we end up with ln(10) · n/a^(n+1) no?

[civic ember]

another important tool in your arsenal is continuity

[fickle granite]

The limit to evaluate is exactly that

[civic ember]

lim n ln (1+1/n) = 1 because lim (1+1/n)^n = e

[fickle granite]

(while the quantities aren't the same, you are allowed to simplify due to continuity of multiplication/composition laws or whatever)

So yes, the problem is basically reduced to checking the limit of n/a^(n+1)

1. does it exist?
2. is it finite?
3. then what is the limit you are trying to evaluate?

[lost badger]

oh and again, so now in this case

0<a<1, infinite (a^(n+1) converges to zero, inf/0 = inf)
a=1, does it matter that the denominator a^(n+1) = 1^(n+1), for the lim n->inf, this wouldn't be indeterminate, correct?

[civic ember]

try to use comparison tests

[lost badger]

since a=1 and not just 1+something that tends to zero

[civic ember]

instead of appealing to some inf/0 arguments and whatnot

$$ \frac{n}{a^{n+1}} \geqslant n $$

[echo tapir]

aL

[civic ember]

if 0<a<1

and since the right side tends to inf

the left side must also have the same limit

[fickle granite]

aL do you know if Stolz-Cesaro can be applied if the limit that is found is infinite?

I can't find it on wikipedia

[lost badger]

afaik stolz-cesaro is appliable as long as the denominator b_n in its limit n->inf is = inf

hence why i did this first

at least in my notes it's a condition for s-c to be appliable

[civic ember]

iirc it is proved by second comparison

which assumes the limit of the ratio is finite

[lost badger]

this is exactly what is in my notes

[fickle granite]

Hmm, ok

I shall trust

[lost badger]

so then, for 0<a<1, since n/a^(n+1) >= n, the limit n->inf (n/a^(n+1)) will be infinite

for a=1, it's still infinite since in this case 1^(n+1) = 1?

or am i assumming incorrectly

[civic ember]

yes, because n / a^{n+1} = n

for every n

[lost badger]

okay, so to clarify since i'm still confused carrying over the fact that "1^inf is an indeterminate form" from high school

[civic ember]

it is correct that it is indeterminate

in the sense that it can be anything

but 1^inf is not valid arithmetic

[lost badger]

my question here is as in, since in this limit it's a=1, and not ie a=1+1/n, it shouldn't be considered an indeterminate form?

that's the way i'm visualising it here

[fickle granite]

It is not indeterminate, but the question revolves about how you justify it

How would you do it?

[lost badger]

i mean i don't really know the way to justify it for a=1 if 1^inf is indeed an indeterminate form

[civic ember]

stop writing 1^inf please

[fickle granite]

a^(n+1) is then always equal to 1

and a sequence that is always equal to 1 (i.e. all its terms are 1)

converges to 1

period

[lost badger]

this is again, because n never really is infinite

right

[civic ember]

yes, n is a natural number

all natural numbers are finite

but it tends to infinity, meaning it gets arbitrarily large values

but they are all finite

[lost badger]

okay, that was my mistake, interpreting that in the limit it IS infinite and not just approaches

[fickle granite]

A pro tip: using indeterminate form arithmetic is a terrible habit imo

as aL said, avoid it like plague

and quit writing 1^inf

[lost badger]

okay

[fickle granite]

you'll only shoot yourself in the foot on the long run

[lost badger]

so then again, a^n+1 converges to 1

[fickle granite]

correct

[civic ember]

if a=1, yes

[fickle granite]

because it is ALWAYS equal to 1

[lost badger]

okay, thank you so much for clarifying this

[civic ember]

are you familiar with the definition of limit?

[lost badger]

i am, i just have a bad habit from high school of directly plugging in the value the incognita approaches

never really thought of it twice for some reason

[civic ember]

you can still do it in your head, but don't ever use it as justification for something

for example this sequence

[lost badger]

yeah, i'm not sure if for example it's valid justification for applying l'hôpital (which we were taught to)

[civic ember]

$$ \left(1+\frac{1}{n}\right)^n $$

[echo tapir]

aL

[fickle granite]

There are very formal hypotheses to the theorem, which you need to formally verify

[civic ember]

you could think of it in terms of "1^inf" because the base converges to 1 and the power is "inf"

but it's not the way to reason for the convergence/divergence of this sequence

1^inf simply means that the limit could be anything

it is ..well.. "indeterminate"

[lost badger]

okay, thank you

so then, for a>1

[civic ember]

then your sequence is of this form

$$ \frac{n}{e^n} $$

[echo tapir]

aL

[lost badger]

wait, why e^n

[fickle granite]

$\frac{n}{\exp(\alpha n)}$ for some $\alpha$

[echo tapir]

Rion

[fickle granite]

And this, buddy, is a usual limit that you are required to know (also by heart)

[civic ember]

from yesterday i think we had e^n/n is unbounded

therefore its inverse must converge to 0

[lost badger]

wait, but why a^(n+1) ~ e^n though

for a>1 that is

[civic ember]

a*a^n

it's just a constant difference

[fickle granite]

$a^n = \exp\left(\ln(a)n\right)$

[lost badger]

oh wait

[echo tapir]

Rion

[lost badger]

yeah okay i forgot

[fickle granite]

Am I the only guy in this server to write exp all the time

[lost badger]

i was thinking of a different form to solve this i guess

[civic ember]

no

[lost badger]

yeah i'm very used to just solving these sort of limits through l'hôpital

even though it's not required

[civic ember]

i mostly write exp(f(x)) too

becareful with lhopital

[lost badger]

is it not appliable here? since you do get an indeterminate form inf/inf

[civic ember]

it is applicable but you might not understand why

[fickle granite]

It is applicable, but...

yeah

[civic ember]

remember that lhopital is defined in case of differentiable functions and so on..

[lost badger]

yes, i mean, i did think of it because you can do the derivative of both the numerator and the denominator

[fickle granite]

Though in our specific case, it is a usual limit you are expected to know ever since high school (at least in the French curriculum)

[civic ember]

yes but your numerator and denominator are not differentiable functions

they are sequences

[lost badger]

oh

i was just interpreting them as you would do for limit of a function

[civic ember]

and you would be right

[fickle granite]

You could, but justification is required

[civic ember]

but you must understand why it won't change the limit

and it because of..

continuity

[civic ember]

.

[lost badger]

i'm struggling to understand the concept of continuity applied here

[civic ember]

well im not sure if im supposed to speak about it then

i don't know where your analysis course is at

[lost badger]

it's not quite an "analysis" course but just a regular calculus course

[civic ember]

but if it is a real analysis course, you must talk about continuity sooner or later

[lost badger]

except the syllabus is kinda odd

i could show you if that even matters

[fickle granite]

Let's say that you prove:
$\frac{\exp(\alpha x)}{x^\beta} \to \infty$ when $x \to \infty$. What makes you say that the statement will still hold when you replace $x$ with $n$ ?

[echo tapir]

Rion

[fickle granite]

An even harder question: say you prove that $\frac{\exp\left(\alpha n \right)}{n^\beta} \to \infty$ when $n \to \infty$. What makes you say that this will still hold when you replace $n$ with $x$ ?

[echo tapir]

Rion

[lost badger]

i can't get to prove why or why not tbh

[lost badger]

could I ask where does this stem from?

[fickle granite]

= exp(ln(a))^n

[lost badger]

ohh it's (ln(a))(n)

[lost badger]

so then, again, from this, how do you justify that this is 0?

i mean i can guess it i just can't think of a way to prove it

[lost badger]

so...

[fickle granite]

Sorry for poofing, I am busy closing all old posts manually

[lost badger]

i understand, im just wondering how to finalise this

[fickle granite]

The limit is zero, this is a common limit. I think it would be wiser to look for a proof online

because this is a really general result that you will use time and time again

I will give you a proof sketch

[lost badger]

for the sake of simplicity could i just use l'hopital here instead?

[fickle granite]

Yes, you can

[lost badger]

since i'm not gonna be asked to prove why is l'hopital appliable in this situation

appliable? applicable? idk

whatever but yes, 1/e^n which is in fact 0 for n->inf

the sole way i could figure out regardless is the growth of the function but since it's a sequence i'm not sure this ever applies

[fickle granite]

In the meantime, I will give you the proof of a first common result:

For any $m$ integer, $\lim_{x \to \infty} \frac{e^x}{x^m} = 0^$

Proof sketch is as follows:

1. We prove that for any $t$, we have $e^t \geq 1+t$

2. Prove that $\exp(x)^\frac{1}{m+1} = \exp\left(\frac{x}{m+1}\right) \geq 1 + \frac{x}{m+1} > \frac{x}{m+1}$

3. Both sides to the power $m+1$. What happens when $x \to \infty$?

[echo tapir]

Rion
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[fickle granite]

you little 💩

Never mind I can't be bothered to fix my latex

[lost badger]

alright and also

i assume by continuity this has to be related to if a function f / f(n) = a_n where a_n is the general term of a sequence, lim n->inf a_n = lim n->inf f(x)

this is what i have in my lecture notes from our teacher as well

[fickle granite]

Sorry for the late answer. Here are a couple of things to consider.

Punctually, a function $f$ is continuous at $x$ if and only if for every sequence $(x_n)$ that converges to $x$, we also have $f(x_n) \to_{n \to \infty} f(x)$

[echo tapir]

Rion

[lost badger]

on a different note, but i think similar to the concepts i was asking about yesterday

is this correct to do? similarly to how you could approximate H(n) ~ ln(n)

for this series i thought the only easy way to solve this is by applying Stirling's approximation and cancelling out the n^n terms, then rewriting the series as ```latex
$\sum_{n=1}^{\infty} \sqrt{2\pi n} (\frac {5}{e})^{n}$

[echo tapir]

lucho

[lost badger]

and my reasoning here was, given 5/e > 1, then (5/e)^n is a strictly increasing factor multiplied by another strictly increasing factor sqrt(2pin) would mean this series diverges, correct?

obviously first proving each term is indeed strictly increasing

or is this reasoning not sufficient?

[fickle granite]

The arrow here makes nonsense

[fickle granite]

Indeed

[fickle granite]

What convergence test?

[lost badger]

bad habit yet again, i apologise

[lost badger]

wait, wdym by this? i mean that if simply proving that those two terms are strictly increasing, this would mean the series is divergent

is that sufficient?

[fickle granite]

No, you established the divergence of another series

How do you justify that the initial series will also diverge?

I do not see it in what you have written

The idea is there but the writing needs to be better to have everything flow like it should

[lost badger]

oh is this the

i'm not sure if i'm correct but

testing for a_n+1/a_n, if >1 for the limit this means it'll diverge?

off the top of my head i'm not sure

[fickle granite]

Let me give you the proof sketch

[lost badger]

i remember seeing this in my notes or something of the like

[fickle granite]

1. a_n ~ b_n, and both are positive, thus their respective series have the same nature. That is, the series are either both convergent or both divergent
2. The series of b_n diverges (this is the only thing you proved so far)
3. Conclude

Does this seem logical to you?

Sorry if I sound a bit too strict on the writing, but mastering this will significantly improve your reasoning skills, so it is important to practice

[lost badger]

i'm not sure, as far as i know i'd have to prove {a_n} ~ {b_n} first?

which i didn't quite do either but i'm not sure if i'm required to

[fickle granite]

You need to

[lost badger]

this is by doing lim n->inf {an}/{bn}, correct?

and if this is = 1, then they are equivalent

[fickle granite]

I mean you do say you use Stirling's approximation, but you never assert that a_n is equivalent to b_n

See what I am getting at?

[lost badger]

yes i mean

so essentially after reaching that final expression for b_n

[lost badger]

i'd need to do this?

and therefore prove they are indeed equivalent

[fickle granite]

Well verify it yes

You already know that n! is equivalent to whatever the Stirling approximation gives you

So you just multiply that by 5^n / n^n

That will suffice to show that n! 5^n /n^n ~ Stirling(n) 5^n / n^n

[lost badger]

i should also just in case note stirling's approximation

[lost badger]

okay, so if i understood it right

i should just write the approximation and multiply times (5/n)^n?

[fickle granite]

Equivalents cannot be added, but they can be multiplied

Yes

You should have seen this in your course about sequences and equivalents, I believe

[lost badger]

yeah the substitution principle afaik

[fickle granite]

Multiplying is a valid arithmetic for equivalents

So you go ahead and do that

It is a valid proof to get to a_n ~ b_n

[lost badger]

alright i'll write it and would you mind correcting the way i expressed it then?

[fickle granite]

Yes, but it is soon lunch time. I will read after lunch

[lost badger]

no problem

[fickle granite]

Ping me if I have not answered in 3 hours

[lost badger]

my question is though, the substitution principle applies to the limits of the multiplication of two sequences

is this then valid for the series too?

well i assume yes since then the respective series for (a_n·c_n) ~ (b_n·c_n) will then be of the same nature

[lost badger]

i could be more rigorous now that i reread this (ie instead of "terms" use "sequences") and maybe note "for a sequence that diverges, for lim n->inf a_n = inf (the limit of its nth term dne/is infinite) then the series of the sequence will also diverge..."

and in general im not 100% sure it's correct

and since ancn ~ bncn, then the series of ancn will also diverge

well and probably proving that they're unbounded above is overkill

[fickle granite]

Assumption: a_n, b_n, c_n are nonzero and a_n ~ b_n.

Conclusion: a_n c_n ~ b_n c_n

Now, from there, if a_n c_n > 0 and b_n c_n > 0, their series have the same nature

Two series aren't equivalent

[lost badger]

oops

[fickle granite]

Maybe the term "series" is ambiguous but the point is... a series is a limit of a partial sum

So a series is in fact a number

The series is said to be divergent when this number is infinity, or the partial sum does not have a limit

And convergent when the number is finite

So a series is not a sequence, you cannot say two series are equivalent

[lost badger]

okay, the equivalence is my mistake

but just by proving that their respective sequences (ancn, bncn) are both divergent should be enough, right?

[fickle granite]

But come to think of it, it suffices to show that a_n c_n is unbounded

[lost badger]

well, you don't even need to prove ancn diverges if you just prove the equivalence between these two

[fickle granite]

You are correct

If a_n c_n is unbounded, then the series of a_n c_n will diverge

[lost badger]

in this case the one i'm proving is unbounded is bncn

and i already proved the equivalence between the limits of ancn and bncn

so ancn ~ bncn, unless i'm mistaken

[fickle granite]

Why does b_n c_n unbounded imply a_n c_n unbounded?

[lost badger]

and since ancn ~ bncn, then the series of ancn will also diverge

[fickle granite]

This might not be true

[lost badger]

i'm using this from my notes

not sure if i just understood it wrong

[fickle granite]

For a sequence, if a limit exists and is infinite, the sequence is unbounded

But the sequence being unbounded does not imply a limit exists

For example take (-1)^n n

It is unbounded but you can't really say much about a limit

[lost badger]

wait, i'm not sure what you mean here

[fickle granite]

Your lecture notes suppose the existence of a limit

[lost badger]

.

oh

[fickle granite]

For a_n c_n

[lost badger]

i did not prove that lim n->inf ancn exists you're right

[fickle granite]

Having an infinite limit is not synonymous to unbounded

[lost badger]

i know that

then what would be the appropriate way to showing ancn ~ bncn? i'm a little bit lost here

you do already have an ~ bn

[fickle granite]

So you must show that b_n c_n has an infinite limit

It does not suffice to show that it is unbounded, because your theorem is not yet applicable

Does that make sense?

[lost badger]

it's not yet applicable because i haven't proven lim n->inf ancn exists, i guess

[fickle granite]

You can switch the roles of a_n and b_n here

[lost badger]

they're proven to be the same yes

you can just switch them?

[fickle granite]

You show that limit of b_n c_n exists and is infinite

Then, you conclude that the limit of a_n c_n exists, and is infinite too

[lost badger]

will that then suffice for proving ancn ~ bncn?

[fickle granite]

if a_n ~ b_n, it is already true that a_n c_n ~ b_n c_n

So it suffices to establish that a_n ~ b_n (which the Stirling approximation does)

[lost badger]

so i only need to prove a_n ~ b_n... but then that is supposedly true for just applying Stirling's approximation

[fickle granite]

Yes

Correct

[lost badger]

so i unnecessarily riddled around that

😭

[fickle granite]

I just want you to keep a clear mind about proving, so to recap

1. verify that a_n ~ b_n. This is true because of Stirling

2. Show that the limit of b_n c_n exists, and is infinite

3. conclude on the existence and value of the limit of a_n b_n

4. Conclude on the nature of the series of a_n c_n

Does this make sense?

[lost badger]

so you're proving the limit because the limit of the partial sum Sn is just the series

or am i mistaken here

what exactly would you need the limit for otherwise if you've already concluded ancn ~ bncn

[fickle granite]

Well if a sequence doesn't converge to zero, the associated series does not converge

[lost badger]

ohh wait yeah

so you're just proving that it does not converge

[fickle granite]

Yes

It does not converge, so it does not converge to zero

[lost badger]

but if i just show that bncn is indeed unbounded, which means it doesn't converge, is that not enough?

since we've already stated that ancn ~ bncn, if bncn does not converge (it's unbounded above) then ancn is not convergent either?

and if the sequence bncn does not converge, its associated series will not converge; same applies for ancn

[fickle granite]

While it does sound intuitively true, you have no result in your lecture notes that claim so

Your lecture notes only deal with the case where b_n c_n tends to infinity

Which is stronger than unboundedness

[lost badger]

okay, i get that

and is it necessary to develop the limit? or can i just immediately show that in the limit it'll be infinite since 5/e > 1

[fickle granite]

You must explain why the limit of b_n c_n is infinite, because it is part of the hypotheses to verify

Here it is in fact simple to explain

[lost badger]

lim n->inf sqrt(2pin) = inf, lim n->inf (5/e)^n = inf since 5/e > 1, then lim n->inf sqrt(2pin) · (5/e)^n = inf

[fickle granite]

Correct

This suffices as an explanation

[lost badger]

alright

because i couldnt bother with developing the limit like a first time limit student

[fickle granite]

You are allowed to compose limits

[lost badger]

alright

[fickle granite]

This is part of the results you learned in your sequence course

That is supposed to avoid you epsilon headaches

[lost badger]

yeah i dont really want to have to apply the full thing

but regardless, proven now the limit of bncn is indeed infinite, limit of ancn is also infinite

[fickle granite]

Yes, indeed

[lost badger]

then ancn is not convergent, its associated series will also not be convergent

that suffices, right

[fickle granite]

Yes

[lost badger]

okay, thank you

[fickle granite]

You are welcome

[lost badger]

this was in our practice test for our upcoming midterms so if this is the hardest it gets i hope i won't have much deducted if i were to make a mistake of the like

oh and just as a final question

it's very unrelated to this but i just wanna get it off real quick before i have to go if that's fine

during our lecture in differentiation for 2 variables here our teacher equaled h=k, k>0

i don't really understand why is this doable?

better said, why is that done

i have the exercise written down if that's needed

[fickle granite]

I am sorry I can't read for poop

Actually, please close this thread, make a new one and ping me in it

[lost badger]

okay

[fickle granite]

This one is getting too long

[lost badger]

thank you as well

+close

[scenic grotto]

Please thank your Helpers before closing!

Please thank the helpers who assisted you by clicking the buttons below. You can thank each helper only once. Once you're done, click "Close Post" to close this thread.

[scenic grotto]

Thank you for your feedback! Rion has been awarded 1 . They now have 13 . They have 2 daily left for today.