need help with proof

[tame pagoda]

proof that 1/2 + 1/3 + 1/4 . . . .. = 1

[tame pagoda]

i dont know how to start with that proof and i really need help

[mellow blaze]

if you are adding infinitely then $\sum_{n=2}^{\infty}\frac{1}{n}=\infty$

[molten swallow]

basyl

[mellow blaze]

or are you trying to disprove it?

[acoustic grail]

The sum of the infinite series 1/2 + 1/3 + 1/4 + ... can be proven to be equal to 1 using a method known as the partial sum

[mellow blaze]

$\S=\sum_{n=2}^{\infty}\frac{1}{n}=\infty\S_3=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}>1$

[molten swallow]

basyl

[mellow blaze]

clearly the partial sum does not prove that?

the series diverges

[mellow blaze]

$\\text{we can see if a series in the form }\sum_{n}^{\infty}\frac{C}{n^p}\text{ where C and p are}\\text{constants converges or diverges by looking at the value of p}\\\text{if }p>1\text{ the series conerges}\\text{if }p\leq 1\text{ the series diverges}$

[molten swallow]

basyl

[mellow blaze]

here p is equal to 1, so the series diverges

[acoustic grail]

$$\text{Let Sn be the nth partial sum of the series:}$$
$$Sn = \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + ... + \frac{1}{n}$$
$$\text{Then, we can rewrite the sum of the series as:}$$
$$\frac{1}{2} + \frac{1}{3} + \frac{1}{4} + ... + \frac{1}{n} + \frac{1}{(n+1)} - \frac{1}{(n+1)}$$
$$\text{The last term,}-\frac{1}{(n+1)}\text{, is subtracted and added back to the sum}$$
$$\text{in order to group the terms into two parts: the sum of the first}$$
$$\text{n terms, and the last term.}$$
$$\text{Thus, we can rewrite Sn as:}$$
$$Sn = (\frac{1}{2} + \frac{1}{3} + \frac{1}{4} + ... + \frac{1}{n}) + \frac{1}{(n+1)} - \frac{1}{(n+1)}$$

$$= (\frac{1}{2} + \frac{1}{3} + \frac{1}{4} + ... + \frac{1}{n}) + (\frac{1}{(n+1)} - \frac{1}{(n+1)})$$

$$= (\frac{1}{2} + \frac{1}{3} + \frac{1}{4} + ... + \frac{1}{n})$$

$$\text{Notice that the partial sum, Sn, is always less than the actual sum of the series. That is}$$
$$Sn < 1$$
$$\text{However, as n approaches infinity, the value of Sn approaches the}$$
$$\text{sum of the series. Thus, we can write:}$$
$$\lim_{n->\infty}\frac{1}{2} + \frac{1}{3} + \frac{1}{4} + ...$$
$$\text{So, the sum of the infinite series} \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + ... \text{is equal to}$$
$$1$$

[molten swallow]

Alex^

[mellow blaze]

even wolfram alpha agrees??

@tame pagoda the series very much diverges, i don't know what @acoustic grail is talking about??

[mellow blaze]

$\a\leq b\implies\frac{1}{a}\geq\frac{1}{b}\iff a, b>0\\therefore \sum_{n=2}^{\infty}\frac{1}{n}\geq\sum_{n=1}^{\infty}\frac{2^{n-1}}{2^n}=\frac{1}{2}+\frac{1}{4}+\frac{1}{4}+\frac{1}{8}\ ...\\\sum_{n=1}^{\infty}\frac{2^{n-1}}{2^n}=\frac{1}{2}+\frac{1}{4}+\frac{1}{4}+\frac{1}{8}\ ...=\frac{1}{2}+\frac{2}{4}+\frac{4}{8}\ ...=\\=\frac{1}{2}+\frac{1}{2}+\frac{1}{2}\ ... =\infty\\\because\sum_{n=2}^{\infty}\frac{1}{n}\geq\sum_{n=1}^{\infty}\frac{2^{n-1}}{2^n},\sum_{n=2}^{\infty}\frac{1}{n}=\infty$

[molten swallow]

basyl

[mellow blaze]

here is an actual proof (or better said a proof disproving what you wanted proofed)

well it needs a little bit of re arangment to make full sense

but thats the gist of it

i would put 1/2 + 1/4 + 1/4 ... part before the summation of that

[sharp pivot]

life getting sucks
Q: Let a be a positive real number. Let f:RR and g:(a)→ R be
the functions defined by f(x)=sin() and g(x)=- 2log, (√x-√a) log, (ev-ev) Then the value of
lim f(g(x)) is

help me in this

[pastel cedar]

Please don't ask your own questions in other people's 'question-page'. If you want people to help you, it's better that you make your own 'question-page'.

[tame pagoda]

guys i couldn't understand i'm 11th grade

is there any simple proofs ?

i think there is an issue with the question

its just not 1 cant be 1

i think i just got rick rolled

[tame pagoda]

but the number is getting smaller so doesn't that mean its not infinite

[tame pagoda]

i think i got rick rolled man

[mellow blaze]

yes, it is getting smaller and smaller, but there is also an infinite ammount of these infinitley small numbers.

[mellow blaze]

yes, it's hard to prove something that isn't true

[mellow blaze]

and the fractions aren't decreasing quickly enough to ensure that this converges

[mellow blaze]

which part don't you understand? i can try to explain it

[graceful plover]

This is absolutely not a place to troll

[tame pagoda]

yes even the calculator when you put 1/2+1/3+1/4 return a number that is greater than 1

but i couldn't understand how a number is getting smaller and still infinite

its like pi or euler

the number only get more numbers after the point

[tame pagoda]

that looks logical but still cant understand

[modest roost]

Could you elaborate on how you came to the conclusions that

(1) the total sum is 1 before having proven anything? [You just assumed this in the middle of your 'proof']

(2) even if the so-called 'actual sum' were to be 1, how come that the partial sum Sn is less then 1? Can't you see, that (1/2)+(1/3)+(1/4) is already bigger than 1?

[molten swallow]

animeonfire

[rustic condor]

Hello, are you looking for a geometric interpretation method?

[rustic condor]

If so, pls do tell, so we can work on a proof

[pastel cedar]

This is not the same series as the one in the question.

[rustic condor]

oopsie

[tame pagoda]

that looks good

[rustic condor]

sorry i misread ur series

[tame pagoda]

but can you give me some hints for the proof , i want to prove it myself
tho my mathematical skills are not that good

[tame pagoda]

yes notice that

[tame pagoda]

but that series you sent = 1 right ?

[rustic condor]

Hi yes

[tame pagoda]

ok i think i will stop working on my series and work on the series you gave me

[rustic condor]

LOL

[mellow blaze]

can't confirm anything (and really sorry if this is human and hand written), but i have a suspicion that he is using Chat GPT

because tools like Chat GPT will generate bs answers that try to prove something that isn't true

[hardy solstice]

The harmonic series diverges

It is not equal to anything

[modest roost]

Here, I gave it a little go myself

$$\sum_{n=2}^{\infty}\frac{1}{n}=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...$$
$$\sum_{n=2}^{\infty}\frac{1}{n}=\sum_{k=1}^{\infty}\frac{1}{p_{1}^k}+\frac{1}{p_{2}^k}+\frac{1}{p_{3}^k}+\frac{1}{p_{4}^k}+...$$
Where $p_i$ is the $i$-th prime number. $\$
We know that
$$\sum_{l=1}^{\infty}\frac{1}{a^l}=\frac{1}{a-1}$$
Using it on the sum above, we get that
$$\sum_{k=1}^{\infty}\frac{1}{p_{1}^k}+\frac{1}{p_{2}^k}+\frac{1}{p_{3}^k}+\frac{1}{p_{4}^k}+...=\frac{1}{(p_{1}-1)^k}+\frac{1}{(p_{2}-1)^k}+\frac{1}{(p_{3}-1)^k}+\frac{1}{(p_{4}-1)^k}+...$$
Since $p_1$ is defined as the first prime number, in other words, 2, $p_1-1=1$.$\$
Since there is an endless amount of prime numbers, and because we can repeat the process shown above indefinitely with every fraction $\frac{1}{x}<1$, we will end up with an infinite series of adding 1 to itself.

[molten swallow]

animeonfire

[modest roost]

It's super messy writing, but it should hold

[graceful plover]

What is the question exactly ?

[modest roost]

We figured out a while ago, that the question doesn't make sense in the way it's formulated

[graceful plover]

So what did you wrote

that Hn diverge

[modest roost]

I tried to show op why the series is not equal to 1

[graceful plover]

Um this is the most complicated way of doing it

Did you even red a proof of: Hn diverges ?

[modest roost]

Nope, I just gave it a shot to prove it myself. A little adventure late at night if you'd want to give it a name

[graceful plover]

I mean your proof is cool dont get me wrong but didnt op wanted an elementary proof ?

I mean op is in 11th grade according to him

[modest roost]

Welp, I'm 11th grade as well, so...

But yeah, I see your point

Not gonna think about it tonight though, I gotta catch some sleep

[graceful plover]

good night

[tame pagoda]

thank you so much man

[tame pagoda]

yep i should've tried to calculate it with a calculator first

tho calculator proves nothing

[tame pagoda]

i think i have to study more of math

can i use that guys ?
$$\frac{a}{1-s}$$

[molten swallow]

artimum

[tame pagoda]

$$ a = \frac{1}{2} , s = \frac{1}{2}
:: \frac{a}{1-s} = \frac{\frac{1}{2}}{1-\frac{1}{2}} = 1$$

[molten swallow]

artimum

[tame pagoda]

** i use this for the series 1/2 + 1/4 + 1/8 .....

i suck at proofing anything

[mellow blaze]

yes

that is a good way to calculate what a series converges to if it has a common ratio

[modest roost]

Nah, you're good, I'm just planning on doing college and school at the same time starting next year

[tame pagoda]

I really want to be a mathematician but i dont have time to study

We are on the same grade but your higher level than me

[tame pagoda]

Oh , thats cool i was afraid that doesn't consider as a proof

[rustic condor]

most of the time proof is simple

so it tends to feel weird because it seems too good to be true

how far along are you in proving? @tame pagoda

[tame pagoda]

not too much

[knotty cradle]

Not every number can be expressed in that way (1/6 is not an power of a prime). The way to do this is to consider an infinite product, such that it generates the harmonic series (ie, an Euler Product). Consider $\zeta_N(1)=\sum_{n=0}^N\frac1 n=1+\frac1 2 + \frac 1 3+ \frac1 4 + \frac1 5 +\cdots$, then $\frac1 2\zeta_N(1)=\frac1 2\sum_{n=0}^N \frac 1 n=\frac1 2+\frac1 4+\frac1 6+\frac1 8+\frac1 {10}+\cdots $ and we have $\zeta_N(1)-\frac1 2\zeta_N(1)=\left(1-\frac1 2\right)\zeta_N(1)=1+\frac1 3+\frac1 5+\frac1 7+\frac1 9+\cdots$. This leaves behind all the numbers not divisible by 2. Similarly, we can remove all the numbers not divisible by 3 as $\left(1-\frac1 2\right)\left(1-\frac1 3\right)\zeta_N(1)=1+\frac1 5+\frac1 7+\frac1 {11}+\frac1 {13}+\cdots$. In general, we have that $\prod_{p\in\mathbb P_{\leq M}}\left(1-\frac 1 p\right)\zeta_N(1)=1+\frac1 k+\cdots$ where k is a prime greater than $M$. Note that this is approximately 1, and we can say that $\zeta_N(1)= \frac 1 {\displaystyle{\prod_{p\in\mathbb P_{\geq M}}(1-\frac 1 p)}}-\varepsilon}$ for some small $\varepsilon>0$.

More concisely, as this is a product, we can write $\zeta_N(1)=\prod_{p\in\mathbb P_{\geq M}}(1-\frac 1 {1-\frac 1 p}}-\varepsilon$

[molten swallow]

Frogster18
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[modest roost]

I know, but the fact that 1/6 isn't considered by me is irrelevant, as 1/6 (and all other non-power-of-prime numbers) are positive. Adding some positive numbers to the infinite amount of 1's doesn't change the fact that the sum is infinite which was the only thing to prove.

If it were to be the otherway around, that every non-power-of-prime number would be subtracted instead of added to the sum your point is absolutely valid, but in this case it's just not necessary.

[knotty cradle]

There's also a pretty glaring error with your calculations. In general, you can not move around terms inside a sum if it is not absolutely convergent (or in this case, as everything is positive, convergent). Also note that the identity $\sum_{l=1}^\infty\frac1 {a^n}$ is valid specifically for an upper bound of $\infty$. Even if you were able to rearrange the sum to get this, you would have $\sum_{n=1}^{\infty}\sum_{p\in\mathbb{P}}\frac 1 {p^n}=\sum_{p\in\mathbb{P}}\sum_{n=1}^{\infty}\frac 1 {p^n}=\sum_{p\in\mathbb{P}}\frac 1 {p-1}$.

The argument for the divergence of this sum is much more difficult.

[molten swallow]

Frogster18

[knotty cradle]

Showing that that sum does diverge does show that the harmonic sum diverges (though I'm not sure if there are any methods of showing the first diverges without requiring the divergence of the second), but is not the best approach to take.

[sterile hinge]

.close