# Does $\sum_{j=1}^{\infty} {\frac{1}{j^k}}$ have a limit? [duplicate]

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.

## marked as duplicate by Martin R, Cesareo, Lee David Chung Lin, José Carlos Santos calculus StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); May 27 at 14:01

• Where is the index $j$ in the sum? – Arnab Auddy May 26 at 17:38
• @ArnabAuddy Uh, I wanted to use $j,$ but see I had instinctively used $i.$ Corrected now, anyway. Thanks. – Allawonder May 26 at 17:40
• 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. – Julian Mejia May 26 at 17:44
• 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. – Travis May 26 at 17:47
• Possible duplicate of Proving limit of a sum – found instantly with Approach0 – Martin R May 27 at 6:42

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 $$1$$.

• Wow! Thanks. But I feel like Watson usually feels after Holmes has explained some piece of deduction to him. – Allawonder May 26 at 17:52
• @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! – auscrypt May 26 at 18:03
• 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. – Allawonder May 26 at 18:54
• 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) – auscrypt May 26 at 19:02

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)

• The index in the second $\lim,$ isn't it supposed to be $s$? – Allawonder May 26 at 17:58
• yes indeed, thanks for pointing it out. – Alexandros May 26 at 17:59

I feel certain this question must have been asked and answered before, but I can't find it.

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

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: $$\zeta(k) \geq \zeta(k+1) \geq \zeta(k+2)$$.

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:

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

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.