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How can I prove that for $s \in \mathbb{C}$, with real part of $s$ being equal to 1, \begin{equation} \sum_{n=1}^{\infty}\frac{1}{n^{s}} \end{equation} diverges?

Thanks a lot!

Ivanko
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    It only diverges when $s = 1$. The Riemann zeta function has a meromorphic continuation to the entire complex plane with a single pole at $s = 1$. If the series diverged for all $s$ with $\textrm{Re}(s) = 1$, then such a continuation would not be possible. – Ethan Alwaise Dec 13 '16 at 23:25
  • @EthanAlwaise Heh, the problem with ye olde Lacunary series. – Simply Beautiful Art Dec 13 '16 at 23:36
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    @EthanAlwaise not at all – reuns Dec 14 '16 at 13:35
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    @EthanAlwaise an example : $\sum_{n=0}^\infty z^{n} = \frac{1}{1-z}$ but the series on the LHS diverges for every $|z| = 1$ – reuns Dec 14 '16 at 13:52

3 Answers3

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$$\left|n^{-s}-\int_n^{n+1} x^{-s}dx\right| = \left|\int_n^{n+1} \int_n^x s t^{-s-1}dtdx\right| <\int_n^{n+1} \int_n^x |s \, n^{-s-1}|dtdx = \left|\frac{s}{2}n^{-s-1}\right|$$

therefore

$$\left|\sum_{n=1}^{N-1} n^{-s}-\frac{1-N^{1-s}}{s-1}\right| = \left|\sum_{n=1}^{N-1} n^{-s}-\int_n^{n+1} x^{-s}dx\right| < \frac{|s|}{2} \sum_{n=1}^{N-1} |n^{-s-1}|$$ and hence : $\displaystyle\sum_{n=1}^N n^{-s}$ converges as $N \to \infty$ if and only if $\displaystyle\frac{1-N^{1-s}}{s-1}$ converges.

$$\boxed{\text{but for } Re(s) = 1, s \ne 1 : \quad \lim_{N \to \infty} N^{1-s} \quad \text{fails to exist}}$$

reuns
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Using formula $(10)$ from this answer, $$ \zeta(s)=\lim_{n\to\infty}\left[\sum_{k=1}^n\frac1{k^s}-\frac1{1-s}n^{1-s}+\frac12n^{-s}\right] $$ converges for $\mathrm{Re}(s)\gt-1$.

For $s=1+it$, where $t\in\mathbb{R}$, $$ \sum_{k=1}^n\frac1{k^s}=\zeta(s)-\frac1{it}e^{-it\log(n)}+O\!\left(\frac1n\right) $$ Because $\lim\limits_{n\to\infty}e^{-it\log(n)}$ does not exist for $t\ne0$, the series on the left does not converge. In fact, it orbits $\zeta(s)$ at a distance of approximately $\frac1{|1-s|}$.

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robjohn
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Note that the Riemann Zeta function diverges only for $s=1$. For $s\neq 1$, however, the function is convergent.As for divergence at $s=1$, there are many proofs extant, one of which is the Cauchy Condensation Test. The function's definition in the form given by you is valid only for $Re(s)>1$.For other regions, we need to use other representations, one of which, as pointed by DonAntonio in the comments, in series form, for $Re(s)>0$ is using Dirichlet Eta Function.The convergence at $Re(s)=1,s\neq1$ can be proved by observing that $\zeta(s)=\frac{1}{1-2^{1-s}}\sum_\limits{n=1}^{\infty}\frac{(-1)^{n-1}}{n^s}$ converges for $Re(s)>0,s\neq1$. The convergence for $\frac{(-1)^{n-1}}{n^s}$ at $Re(s)=1$ is proved using the theory of convergence of Dirichlet series to the right of the absscica of convergence, which, in this case, is $Re(s)>0$. It may also be seen by expanding $\zeta(s)$ around $s=1$ as a Laurent Series giving $\zeta(s)=\frac{1}{s-1}+\sum_\limits{n=0}^{\infty}\frac{(-1)^n}{n!}\gamma_n(s-1)^n$, where $\gamma_n$ is the Steiljes Constant. The Laurent series can be derived by using the integral representation of Riemann Zeta function and several methods can be found here.

The theorem on convergence of Dirichlet Series $\frac{(-1)^{n-1}}{n^s}$ is proved using boundedness of partial sums of $|(-1)^{n-1}|\leq1$ and the use of Abel's summation Formula or Euler's Summation Formula on $\frac{(-1)^{n-1}}{n^s}$. Alternatively, a proof of the same is given in Robjon's answer here using Generalized Dirichlet's test for sum of product of two sequences.

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vidyarthi
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