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Does $\sum_{k=1}^\infty\frac1{k^n}$ converge for $\Re(n)=1,\Im(n)\ne0$?

The ratio test is inconclusive.

It passes the term test for $\Re(n)=1$, but this is not sufficient to prove convergence.

Since we are dealing with so many complex numbers, I do not know of any convergence tests for this. I know that if it is convergent, it is conditionally convergent.

Simply Beautiful Art
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2 Answers2

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In THIS ANSWER, I used the Euler-Maclaurin Summation Formula (EMSF) to show that the series $\sum_{k=1}^\infty \frac{1}{k^n}$ converges for $\text{Re}(n)>1$ and diverges for $\text{Re}(n)<1$.

Proceeding similarly, using the EMSF we can write for $n=1+iy$

$$\begin{align} \sum_{k=1}^N \frac{1}{k^{1+iy}}&=\sum_{k=1}^N \frac{e^{-iy\log(k)}}{k}\\\\ &=\frac{\sin(y\log(N))}{y}+i\frac{\cos(y\log(N))}{y}+C+O\left(\frac1N\right) \end{align}$$

for some constant $C$.

Since both $\lim_{N\to \infty}\sin(y\log(N))$ and $\lim_{N\to \infty}\cos(y\log(N))$ diverge for $y\ne 0$, the series of interest diverges likewise.

Community
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Mark Viola
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$$n^{-s} - \int_n^{n+1} x^{-s} dx = \int_n^{n+1} (n^{-s}-x^{-s}) dx = \int_n^{n+1} \int_n^x s y^{-s-1}dy dx = \mathcal{O}(s n^{-s-1})$$ so that $$\begin{eqnarray}\sum_{n=1}^\infty (n^{-s}-\int_n^{n+1} x^{-s} dx) &=& \lim_{N \to \infty} \sum_{n=1}^N n^{-s}- \int_1^{N+1}x^{-s}dx \\ &=& \lim_{N \to \infty} \sum_{n=1}^N n^{-s} + \frac{(N+1)^{1-s}-1}{s-1}\underset{\qquad \scriptstyle (s \ne 1)}{}\end{eqnarray}$$ converges for $Re(s) > 0$.

Since $N^{1-s}$ diverges whenever $Re(s) \le 1$, so does $\sum_{n=1}^N n^{-s}$.

Note that by analytic continuation : $$\zeta(s) = \frac{1}{s-1}+ \sum_{n=1}^\infty (n^{-s} - \int_n^{n+1} x^{-s} dx)\qquad \quad (Re(s) > 0)$$

reuns
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  • Are you saying that $\sum n^{-s}$ converges if $\sum n^{-s}-\int_n^{n+1}x^{-s}dx$ converges? I'm not sure, but at the very least, I know that $\sum_{n=1}^\infty\int_n^{n+1}x^{-s}dx$ diverges. Just confused on why we got to that integral. – Simply Beautiful Art Aug 31 '16 at 18:45
  • @SimpleArt no it is not what I said – reuns Aug 31 '16 at 18:49
  • Then how does it relate to the divergence/convergence of $\sum n^{-s}$? – Simply Beautiful Art Aug 31 '16 at 18:50
  • I already know how to extend the Riemann zeta function to pretty much any value, but I'm just interested in that summation ^^ – Simply Beautiful Art Aug 31 '16 at 18:51
  • @SimpleArt honestly I wrote everything – reuns Aug 31 '16 at 18:51
  • Oh, so it is clear that $\sum n^{-s}-\int x^{-s}dx$ converges regardless of $s$, so it is merely a matter of whether $\int_1^{N+1}x^{-s}dx$ converges. Did I get that right? And thank you. – Simply Beautiful Art Aug 31 '16 at 18:56
  • @SimpleArt your notations are horrible :) mines are correct : as $N \to \infty$, $\ \ (\sum_{n=1}^N n^{-s}) - \int_1^{N+1} x^{-s}dx$ converges for $Re(s) >0$ – reuns Aug 31 '16 at 18:57
  • Sh... just short-handing. These are the comments, and you get what I mean. – Simply Beautiful Art Aug 31 '16 at 18:58
  • @SimpleArt now, do you understand the last line ? – reuns Aug 31 '16 at 19:16
  • Yeah, I think so. – Simply Beautiful Art Aug 31 '16 at 19:17
  • @SimpleArt ok explain it – reuns Aug 31 '16 at 19:19
  • Well, since $\lim_{N\to\infty}\sum_{n=1}^N(n^{-s}-\int_n^{n+1}x^{-s}dx)$ converges for $\Re(s)>0$ and is equal to $\lim_{N\to\infty}\sum_{n=1}^Nn^{-s}-\int_1^{N+1}x^{-s}dx$, $\sum_{n=1}^\infty n^{-s}$ converges if $\int_1^\infty x^{-s}dx$ converges, and the integral only converges for $\Re(s)>1$. Note you can use \Re – Simply Beautiful Art Aug 31 '16 at 19:23
  • @SimpleArt I meant the last line with the analytic continuation of $\zeta(s)$ for $Re(s) > 0$ – reuns Aug 31 '16 at 19:25
  • Oh, but that's quite frankly obvious, since the integral is equal to $1/(s-1)$ if it converges. :/ – Simply Beautiful Art Aug 31 '16 at 19:26
  • @SimpleArt No. What is obvious is that $F(s) = \sum_{n=1}^\infty (n^{-s} - \int_n^{n+1} x^{-s}dx)$ is analytic on $Re(s) > 0$ (why ?), and since for $Re(s) > 1$ we have $(s-1)F(s) = (s-1)\zeta(s) - 1$, by analytic continuation this stays true for $Re(s) > 0$. – reuns Aug 31 '16 at 19:30
  • (note that I didn't define $\zeta(s)$ for $Re(s) \in (0,1]$ yet, I just supposed it had an analytic continuation, that indeed I just proved. of course, using $\eta(s) = (1-2^{1-s})\zeta(s)$, it also proves that $(1-2^{1-s})(F(s)(s-1)+1) = \eta(s)(s-1)$) – reuns Aug 31 '16 at 19:33
  • :/ well, that's not so obvious if you haven't taken Calc. II. Um, yeah. I just know that if $\zeta(s)$ is analytic and follows some functional equation that is analytic on a domain beyond the zeta function's original domain, then the analytic continuation of the zeta function must still follow that functional equation. That's my best definition of analytic continuation so far. – Simply Beautiful Art Aug 31 '16 at 19:35
  • @SimpleArt you should look at $\frac{1}{1-z}$ the analytic continuation of $\sum_{n=0}^\infty z^n$, then at $\Gamma(s)$, and of course at a complex analysis course – reuns Aug 31 '16 at 19:38
  • Working towards that complex analysis course, and already know how to analytically continue the zeta and gamma function with some extremely bad notation and half-complete logic. :/ – Simply Beautiful Art Aug 31 '16 at 19:40