Solve the following limit

[reef belfry]

• (I need to post the picture first but for some reason it's not letting me)

[proper flicker]

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:

[reef belfry]

The exercise itself is to study the convergence for a given series depending on the values of x, x is a real number

But I arrived at this indetermination for |x|>1 and I'm not sure how to solve

I know that the limit will be inf since x^n for |x|>1 will always grow quicker than sqrt(n)

However I need to justify this and I'm not sure how to either

[radiant basin]

$$ \frac{1}{R} = \limsup _n \sqrt[n]{|a_n|} $$

[solid aurora]

aL

[radiant basin]

a_n = 1 / sqrt(n) in your case

R is the radius of convergence

this sort of reasoning is not correct

[reef belfry]

So what's the correct way to justify it

[radiant basin]

calculate the radius of convergence by the formula I gave you

you have a power series of the form

$$ \sum a_nx^n $$

[solid aurora]

aL

[radiant basin]

it's a power series about the point 0

[reef belfry]

I'm pretty bad with sequences/series so I'm not sure how to do all this lol sorry

Could you explain how would you do this if you don't mind

for a_n = 1/sqrt(n)?

[radiant basin]

$$ \lim _n \frac{1}{\sqrt[2n]{n}} $$

what is this limit

[solid aurora]

aL

[reef belfry]

for n->inf?

[radiant basin]

yes

do you know that

$$ \lim \sqrt[n]{n} = 1 $$

[solid aurora]

aL

[reef belfry]

is this not 1/(1/2)

sorry for the late answer

[radiant basin]

the limit is 1

which means your radius of convergence is also 1

what that means is the series converges absolutely for any x in (-1,1)

you have to check separately whether you have convergence for x=1 and x=-1

[reef belfry]

are you clearing for R here

I'm confused since I haven't seen this formula

[twin spruce]

Lucho what happened to the previous thread

[radiant basin]

this problem is about convergence of power series

[twin spruce]

aL start from the absolute basics

[radiant basin]

and the radius of convergence formula has to be one of the first things you see

[reef belfry]

well not quite

but you are clearing for R

[radiant basin]

R=1 as per the formula

[twin spruce]

what note taking app do you use btw

[reef belfry]

sounds dumb but if I saw this solved step by step I'd probably understand better lol

[reef belfry]

goodnotes

[twin spruce]

appreciate it 👍

[radiant basin]

we are solving it step by step

[reef belfry]

not really if you jumped from the formula straight to the result

[twin spruce]

you mean looking at a solution and then explaining it to you?

[reef belfry]

just the solution itself step by step

[twin spruce]

that's what aL is doing here

infact he is taking you through the steps

[radiant basin]

1. Determine at which point the power series is defined
2. Calculate its radius of convergence
3. If R>0 check the end points of the interval for convergence

[reef belfry]

because i don't get how is the limit written for the convergence radius formula

[twin spruce]

Let me know if I misinterpreted, do you need help understanding why/what's the proof of R being the radius of convergence?

[reef belfry]

no

I mean how do you write the limit itself

the a_n in here is of form 1/sqrt(n)?

[radiant basin]

yes

you posted this under real analysis tag, so I have to assume you have covered power series in your lectures

[reef belfry]

nvm i was just tweaking on how to write the limit itself lol

[radiant basin]

ok, are you up to speed now?

[reef belfry]

so R = 1, then what else is the justification again

[reef belfry]

and unfortunately no

[radiant basin]

at which point is the power series defined?

[reef belfry]

wait so this is just applying cauchy-hadamard theorem

[radiant basin]

good

[reef belfry]

given radius of convergence 1 don't you then check for |x| > 1 and |x|<1 like i did?

[radiant basin]

no

it's already clear that |x|<1 is the case where absolute convergence occurs by the C-H theorem

but the theorem doesn't say anything about endpoints of the interval

[reef belfry]

btw if I'm taking time to respond it's because i have to find the translation to the english terms since i do all my math in spanish

sorry

[radiant basin]

not a problem at all

I will know what you're talking about

[reef belfry]

no because I can't find the translation for endpoints

[radiant basin]

(-1,1) is the open interval with endpoints -1 and 1

[reef belfry]

do you mean whether {1} and {-1} are included

[radiant basin]

the C-H thm implies that this power series converges absolutely for every x in (-1,1)

we don't need to worry about those

but the radius itself is 1, so we have to check for convergence at x=1 and x=-1 separately

[reef belfry]

yes

that's what I meant

does this not cause an indetermination 1^inf in the numerator for lim n->inf

[radiant basin]

well, find out

for x=-1 you have

$$ \sum \frac{1}{\sqrt{n}}(-1)^n $$

[solid aurora]

aL

[radiant basin]

this converges, yes?

[reef belfry]

you have an indetermination...?

unless I'm confused

[radiant basin]

nope

this is a convergent series

[reef belfry]

1/sqrt(n) for n->inf converges to 0

does the (-1)^inf not influence

[radiant basin]

there is no (-1)^înf

[reef belfry]

?

[radiant basin]

what do you know about convergence of alternating series?

serie alterna

theorema de leibniz

[reef belfry]

for {a_n} monotonically decreasing and since lim n->inf = 0 the series converges

[radiant basin]

muy bien 🙂

no mention (-1)^inf or what have you

[reef belfry]

1/sqrt(n) monotonically decreasing so the series converges for x=-1 no?

[radiant basin]

that's correct

now what about x=1

[reef belfry]

for x=1 you can't rewrite it like this can you

i mean

you could

[radiant basin]

$$ \sum \frac{1}{\sqrt{n}} $$

[solid aurora]

aL

[radiant basin]

la serie armonica

[reef belfry]

i'm confused

you have 1^n/sqrt(n)

[radiant basin]

and 1^n = 1 for every n, yes?

[reef belfry]

only for n is a real number

[radiant basin]

natural number even

and it is

[reef belfry]

but in the series n=1 to inf

[radiant basin]

n never becomes infinity

it's always finite

but it tends to infinity

[reef belfry]

are you never supposed to assume n becomes infinity in a series

[radiant basin]

of course

[reef belfry]

that's something i was not taught in my lectures, thank you

[radiant basin]

$$ \sum _{n=1}^\infty a_n = \lim _{m\to\infty} \sum _{n=1}^ma_n $$

[solid aurora]

aL

[radiant basin]

this is the definition

[reef belfry]

so then it's for x=1 it converges too?

or does 1/sqrt(n) not converge?

not sure how the criteria goes for this lol

[radiant basin]

diverges

$$ \sum \frac{1}{\sqrt{n}} \geqslant \sum \frac{1}{n} = \infty $$

[solid aurora]

aL

[radiant basin]

es una serie armonica

[reef belfry]

yeah i've not done the integral test for this i have no idea what the reasoning is

[radiant basin]

well, you must have covered this by now, I refuse to believe you didn't

[reef belfry]

i mean i assume they did i just missed the first few lectures due to illness

[radiant basin]

but you do have lecture notes for the course, yes?

[reef belfry]

i just bluntly assumed it diverges

yes, but they're very surface level and don't dig too deep into reasoning

[radiant basin]

oh brother 😦

[reef belfry]

Ohh is this the comparison criteria

yeah okay sorry

[radiant basin]

very good

so all in all, the power series converges absolutely in (-1,1), converges in [-1,1)

and diverges everywhere else

[reef belfry]

so to sum it all up, calculate the radius of convergence (according to cauchy theorem) and evaluate for the endpoint values

[radiant basin]

that is correct

$$ \sum a_n (x-2)^n $$

[solid aurora]

aL

[radiant basin]

where is this power series defined?

[reef belfry]

what again is the difference between converges absolutely in (-1,1) and converges in [-1,1)?

also i'm just taking long to process because language barrier

defined means what here

[radiant basin]

the series

$$ \sum a_nx^n $$

[solid aurora]

aL

[radiant basin]

is defined at point 0 becuase it's of the form (x-0)^n

[reef belfry]

i seriously have a problem with understanding math in english

[radiant basin]

la serie de potencias?

[reef belfry]

analogically for this it'll be defined in point x=2?

what

[radiant basin]

correct!

[reef belfry]

oh so it'll be defined for any x^n = 1

orrrrr

i'm confused

[radiant basin]

by definition

$$ \sum a_n(x-a)^n $$

[solid aurora]

aL

[radiant basin]

is a power series defined at x=a

[reef belfry]

oh i remember now

i apologise

[radiant basin]

now suppose your power series is defined at x=2

and has radius of convergence 3

in which interval does the series converge absolutely?

[reef belfry]

if i were to give an answer it'd have no basis behind it i'm just confused

[radiant basin]

you already know everything necessary to answer it

[reef belfry]

im just guessing (-3,3) since it's the radius of convergence

[radiant basin]

the key element is the C-H theorem

[radiant basin]

but not around 0

but around 2

[reef belfry]

oh

(-1,5) ?

wait

[radiant basin]

correct

[reef belfry]

oh okay

thank you

do i need to close this thread or? i have to rewrite it once i get home

[radiant basin]

you can +close it if you think that's all, yes

[reef belfry]

Can I also use this same channel for a different question?

[radiant basin]

close this one and make a new question

easier for others to follow

@reef belfry

[reef belfry]

okay thanks

+close

[near meadow]

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.