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This question already has an answer here:

I was playing with the calculator recently and I noticed that we seem to have, as $k\to\infty,$ the limiting value of $$\sum_{j=1}^{\infty} {\frac{1}{j^k}}$$ to be $1.$

Is this in fact true? If so how can one see it? In particular, is there a proof for this somewhere?

Thanks.

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marked as duplicate by Martin R, Cesareo, Lee David Chung Lin, José Carlos Santos calculus May 27 at 14:01

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ Where is the index $j$ in the sum? $\endgroup$ – Arnab Auddy May 26 at 17:38
  • $\begingroup$ @ArnabAuddy Uh, I wanted to use $j,$ but see I had instinctively used $i.$ Corrected now, anyway. Thanks. $\endgroup$ – Allawonder May 26 at 17:40
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    $\begingroup$ So, a way to rephrase your question is that you want to prove that $\lim_{k\to\infty} \zeta(k)=1$ This is in fact true. $\endgroup$ – Julian Mejia May 26 at 17:44
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    $\begingroup$ The value of this series (where it converges) coincides with value of $\zeta(k)$, where $\zeta$ is the Riemann zeta function: en.wikipedia.org/wiki/Riemann_zeta_function. $\endgroup$ – Travis May 26 at 17:47
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    $\begingroup$ Possible duplicate of Proving limit of a sum – found instantly with Approach0 $\endgroup$ – Martin R May 27 at 6:42
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Note that $$\sum_{j=2}^\infty j^{-k} < \int _1 ^\infty x^{-k} \mathrm{d}x$$ (the left hand side is the lower sum of rectangles). The RHS evaluates to $\frac{1}{k-1}$ which clearly tends to $0$ as $k \rightarrow \infty$, and adding back in the $j=1$ term gives us a limit of $1$.

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  • $\begingroup$ Wow! Thanks. But I feel like Watson usually feels after Holmes has explained some piece of deduction to him. $\endgroup$ – Allawonder May 26 at 17:52
  • $\begingroup$ @Allawonder The proof was inspired from the integral test for convergence (if you haven’t heard of that before, see en.m.wikipedia.org/wiki/Integral_test_for_convergence). Hopefully that helps explain the motivation behind the solution! $\endgroup$ – auscrypt May 26 at 18:03
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    $\begingroup$ Oh no, you misunderstood me. Apparently you don't know about Holmes and Watson. In any case I was referring to the fact that it seems so obvious after the fact, that I felt foolish for having asked in the first place instead of thinking a little bit more. $\endgroup$ – Allawonder May 26 at 18:54
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    $\begingroup$ Oh.. oh... That’s embarrassing :( I’m gonna go hide under a table ┬─┬ (you can't see me in the diagram because I'm hiding) $\endgroup$ – auscrypt May 26 at 19:02
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we have that $ζ(s)$ converges absolutely to the right of $\operatorname{Re}(s)=1$, therefore:

$\lim_{s \to \infty}ζ(s)=\sum_{j=1}^\infty \left(\lim_{s\to\infty}j^{-s}\right)=1+0+...=1$, as you correctly pointed out.

(here you put $\lim$ inside the sum)

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  • $\begingroup$ The index in the second $\lim,$ isn't it supposed to be $s$? $\endgroup$ – Allawonder May 26 at 17:58
  • $\begingroup$ yes indeed, thanks for pointing it out. $\endgroup$ – Alexandros May 26 at 17:59
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I feel certain this question must have been asked and answered before, but I can't find it.

Consider that$$ 1+2^{-j}+3^{-j}+\cdots<1+(2^{-j}+2^{-j})+(4^{-j}+4^{-j}+4^{-j}+4^{-j})+\cdots$$

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Note that your sum is simply $\zeta(k)$. We have

$1 \leq \sum_{j=1}^{\infty} {\frac{1}{j^{2k}}}= \zeta(2k)= (-1)^{k+1} \frac{B_{2k} (2 \pi)^{2k}}{2 (2k)!}$ (see for example https://en.wikipedia.org/wiki/Particular_values_of_the_Riemann_zeta_function).

This goes to 1 for $k \rightarrow \infty$, now use monotony: $ \zeta(k) \geq \zeta(k+1) \geq \zeta(k+2)$.

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Just to give a simple, relatively self-contained proof, note that for $j\gt1$, we have

$${1\over j^k}\lt{1\over kj^2}$$

if $k\gt4$, since $j^{k-2}\ge2^{k-2}\gt k$ for $k\gt4$. It follows that for large values of $k$ we have

$$1+{1\over2^k}+{1\over3^k}+\cdots\lt1+{1\over k}\left({1\over2^2}+{1\over3^2}+\cdots\right)=1+{1\over k}\left({\pi^2\over6}-1\right)\to1$$

Remark: It's enough to know that ${1\over2^2}+{1\over3^2}+\cdots$ converges; its exact value is incidental. We could make things even more self-contained by further weakening the key inequality to

$${1\over j^k}\lt{1\over kj^2}\lt{1\over kj(j-1)}={1\over k}\left({1\over j-1}-{1\over j}\right)$$

so that we get a telescoping series in the upper bound:

$$1+{1\over2^k}+{1\over3^k}+\cdots\lt1+{1\over k}\left(\left({1\over1}-{1\over 2}\right)+\left({1\over2}-{1\over 3}\right)+\left({1\over3}-{1\over 4}\right)+\cdots \right)=1+{1\over k}\to1$$

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Others have already addressed the fact that your sum converges for $k>1$. It's also clear it reduces as $k$ increases, since each individual term with $j\ne 1$ reduces. But what is the $k\to\infty$ limit?

Before we start, $\delta_{jl}$ is shorthand for $1$ if $j=l$ and $0$ otherwise. Obviously $\lim_{k\to\infty}j^{-k}=\delta_{j1}$ for integers $j\ge 1$, because if $j=1$ then $j^{-k}=1$ for all $k$, and if $j>1$ then as $k\to\infty$ the function $j^{-k}$ exponentially decays to $0$. Thus$$\lim_{k\to\infty}\sum_{j\ge 1}j^{-k}=\sum_{j\ge 1}\lim_{k\to\infty}j^{-k}=\sum_{j\ge 1}\delta_{j1}=1+0+0+\cdots=1.$$The sum-limit exchange uses dominated convergence.

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