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I now want to know if the series $$\sum_{n=1}^\infty \frac{1}{n^s},$$ where $s$ is a complex number, diverges when $Re(s) \le 1$.

I am not working with analytic continuation. I am working with only the simple definition above.

Is it true that this series diverges for $\Re(s) \le 1$? How can I prove it?

Math Learner
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    It does not work like this. The formula for $\zeta(s)$ you listed is valid for $Re(s) > 1$. Then one proofs that the function can be extended in unique way to holomorphic functions on $Re(s) > 0$ except one point $1$. Nobody claims that the same formula holds in every case. – Salcio Jan 02 '23 at 17:26
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    Lone Learner, I've edited your post in a way that I hope reflects your intention; please correct me if I'm wrong. – Greg Martin Jan 02 '23 at 17:33
  • For $\Re s <1$ this follows by general theory since convergence at any point would imply convergence on the right half plane of that and the series is obviously divergent at $1$; for $\Re s =1, s\ne 1$ the result is trickier and requires some theory – Conrad Jan 02 '23 at 17:34
  • The partial sums are actually bounded for $\Re s =1, s\ne 1$ – Conrad Jan 02 '23 at 17:44
  • @GregMartin Thank you for the edit! Yes it reflects my intention. I have edited it further and removed all mentions of $\zeta(s)$ now because anytime I say $\zeta(s)$ the discussion goes to analytic continuation which I am trying to avoid here. I am trying to learn the basics first before I move on to analytic continuation. – Math Learner Jan 02 '23 at 18:38
  • @Conrad Thank you for explaining the divergence of the series when $\Re(s) < 1$. Solves half of the question for me. So only thing left to show now is that it diverges for $\Re(s) = 1$. – Math Learner Jan 02 '23 at 18:44
  • For that you need to show the partial sum formula $\zeta(1+it)-\sum_{n \le N}n^{-1-it}=-\frac{N^{-it}}{it}+O(N^{-1})$ for fixed $t \ne 0$ and $N > 100t$ say; letting $N \to \infty$ shows the oscillation of the series – Conrad Jan 02 '23 at 19:09
  • It is true that $$\sum_{1\le n\le N}n^{-s}={N^{1-s}\over1-s}+\zeta(s)+O(|s|N^{-\Re(s)})$$, where $\zeta(s)$ is the analytically continued version of the series. – TravorLZH Jan 03 '23 at 19:42

3 Answers3

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Probably the most frequently cited proof of the divergence of $\sum_{{n} = {1}}^{\infty} \frac{1}{n^{s}}$ for $\Re\left( s \right) \leq 0$ is probably the one via the n.th-term test for divergence, which states "If $\lim _{{n} \to {\infty}} a_{n} \neq 0$ or if the limit does not exist, then $\sum _{{n} = {1}}^{\infty} a_{n}$ diverges.".

for $\Re\left( s \right) \leq 0$: $$ \begin{align*} \sum_{{n} = {1}}^{\infty} \frac{1}{n^{s}} &= \sum_{{n} = {1}}^{\infty} \frac{1}{n^{\Re\left( s \right) + \Im\left( s \right) \cdot \mathrm{i}}}\\ \sum_{{n} = {1}}^{\infty} \frac{1}{n^{s}} &= \sum_{{n} = {1}}^{\infty} \frac{1}{n^{\Re\left( s \right)} \cdot n^{\Im\left( s \right) \cdot \mathrm{i}}} &\quad\mid\quad a_{n} = \frac{1}{n^{\Re\left( s \right)} \cdot n^{\Im\left( s \right) \cdot \mathrm{i}}}\\ \sum_{{n} = {1}}^{\infty} \frac{1}{n^{s}} &= \sum_{{n} = {1}}^{\infty} a_{n}\\ \Rightarrow \lim_{{n} \to {\infty}} a_{n} &= \lim_{{n} \to {\infty}} \frac{1}{n^{\Re\left( s \right)} \cdot n^{\Im\left( s \right) \cdot \mathrm{i}}}\\ \lim_{{n} \to {\infty}} a_{n} &= \lim_{{n} \to {\infty}} \frac{1}{n^{\Re\left( s \right)}} \cdot \lim_{{n} \to {\infty}} \frac{1}{n^{\Im\left( s \right) \cdot \mathrm{i}}}\\ \lim_{{n} \to {\infty}} a_{n} &= \left( \lim_{{n} \to {\infty}} \frac{1}{n} \right)^{\Re\left( s \right)} \cdot \lim_{{n} \to {\infty}} \frac{1}{n^{\Im\left( s \right) \cdot \mathrm{i}}}\\ \lim_{{n} \to {\infty}} a_{n} &= \left( 0 \right)^{\Re\left( s \right)} \cdot \lim_{{n} \to {\infty}} \frac{1}{n^{\Im\left( s \right) \cdot \mathrm{i}}}\\ \lim_{{n} \to {\infty}} a_{n} &= \left( 0 \right)^{0} \cdot \lim_{{n} \to {\infty}} \frac{1}{n^{\Im\left( s \right) \cdot \mathrm{i}}} &\Rightarrow \text{no convergence} \tag{if Re(s) = 0}\\ \lim_{{n} \to {\infty}} a_{n} &= \left( 0 \right)^{- \left|\Re\left( s \right) \right|} \cdot \lim_{{n} \to {\infty}} \frac{1}{n^{\Im\left( s \right) \cdot \mathrm{i}}} = \underbrace{\frac{1}{0 ^{\left|\Re\left( s \right) \right|}}}_{= \frac{1}{0}} \cdot \lim_{{n} \to {\infty}} \frac{1}{n^{\Im\left( s \right) \cdot \mathrm{i}}} &\Rightarrow \text{no convergence} \tag{if Re(s) < 0}\\ \end{align*} $$

You could also say $\lim_{{n} \to {\infty}} \frac{1}{n^{\Im\left( s \right) \cdot \mathrm{i}}} = \left( \lim_{{n} \to {\infty}} \frac{1}{n} \right)^{\Im\left( s \right) \cdot \mathrm{i}} = 0^{\Im\left( s \right) \cdot \mathrm{i}}$, and since $0^{\mathrm{i}}$ isn't defined, neither case converges too. However, I find the variant shown more attractive because it does not require a particularly high level of understanding of complex numbers. After all, $\frac{1}{0^{\left| \Re\left( s \right) \right|}} = \frac{1}{0} \to \text{not defined}$ and $\left( 0 \right)^{0} = 0^{0} \to \text{not defined}$ don't have much to do with it.

If you have $0 < \Re\left( s \right) \leq 1$, then arguably the most effective way of showing the divergence of it via the Integral test for convergence aka Maclaurin–Cauchy test. It says "Consider an integer $N$ and a function $f$ defined on the unbounded interval $[N, \infty)$, on which it is monotone decreasing. Then the infinite series $\sum_{n=N}^{\infty} f\left( n \right)$ converges to a real number if and only if the improper integral $\int_{N}^{\infty} f(x) \operatorname{d}x$ is finite. In particular, if the integral diverges, then the series diverges as well.": $$ \begin{align*} \int_{1}^{\infty} \frac{1}{n^{s}} \operatorname{d}n < \sum_{{n} = {1}}^{\infty} \frac{1}{n^{s}} &< 1 +\int_{1}^{\infty} \frac{1}{n^{s}} \operatorname{d}n\\ \int_{1}^{\infty} \frac{1}{n^{s}} \operatorname{d}n &= \int_{1}^{\infty} n^{-s} \operatorname{d}n = \lim_{{x} \to {\infty}} F\left( x \right) - F\left( 1 \right)\\ \end{align*} $$ and thus we get:

for $\Re\left( s \right) = 1$ and $\Im\left( s \right) = 0$: $$ \begin{align*} F\left( n \right) = \int_{0}^{x} n^{-1 + \Im\left( s \right)} \operatorname{d}n &= \ln\left( n \right)\\ \int_{1}^{\infty} \frac{1}{n^{s}} \operatorname{d}n &= \lim_{{x} \to {\infty}} \left( \ln\left( x \right) \right) - \ln\left( 1 \right)\\ \int_{1}^{\infty} \frac{1}{n^{s}} \operatorname{d}n &= \infty - 0\\ \int_{1}^{\infty} \frac{1}{n^{s}} \operatorname{d}n &= \infty \Rightarrow \text{no convergence}\\ &\Rightarrow \infty < \sum_{{n} = {1}}^{\infty} \frac{1}{n^{s}} \Rightarrow \text{no convergence of } \sum_{{n} = {1}}^{\infty} \frac{1}{n^{s}} \tag{if Re(s) = 1 and Im(s) = 0}\\ \end{align*} $$ for $0 < \Re\left( s \right) < 1$: $$ \begin{align*} F\left( n \right) = \int_{0}^{x} n^{-s} \operatorname{d}n &\overbrace{=}^{\text{power rule}} \frac{1}{-s + 1} \cdot n^{-s + 1}\\ \int_{1}^{\infty} \frac{1}{n^{s}} \operatorname{d}n &= \lim_{{x} \to {\infty}} \left( \frac{1}{-s + 1} \cdot x^{-s + 1} \right) - \frac{1}{-s + 1} \cdot 1^{-s + 1}\\ \int_{1}^{\infty} \frac{1}{n^{s}} \operatorname{d}n &= \frac{1}{-s + 1} \cdot \lim_{{x} \to {\infty}} \left( x^{-s + 1} \right) - \frac{1}{-s + 1}\\ \int_{1}^{\infty} \frac{1}{n^{s}} \operatorname{d}n &= \frac{1}{-s + 1} \cdot \lim_{{x} \to {\infty}} \left( x^{-\Re\left( s \right) - \Im\left( s \right) \cdot \mathrm{i} + 1} \right) - \frac{1}{-s + 1}\\ \int_{1}^{\infty} \frac{1}{n^{s}} \operatorname{d}n &= \frac{1}{-s + 1} \cdot \lim_{{x} \to {\infty}} \left( x^{-\Re\left( s \right) + 1} \cdot x^{-\Im\left( s \right) \cdot \mathrm{i}} \right) - \frac{1}{-s + 1}\\ \int_{1}^{\infty} \frac{1}{n^{s}} \operatorname{d}n &= \frac{1}{-s + 1} \cdot \underbrace{\lim_{{x} \to {\infty}} \left( x^{\overbrace{-\Re\left( s \right) + 1}^{1 > \Re\left( s \right)}} \right)}_{\to \infty} \cdot \lim_{{x} \to {\infty}} \left( x^{-\Im\left( s \right) \cdot \mathrm{i}} \right) - \frac{1}{-s + 1}\\ \int_{1}^{\infty} \frac{1}{n^{s}} \operatorname{d}n &= \frac{1}{-s + 1} \cdot \infty \cdot \lim_{{x} \to {\infty}} \left( x^{-\Im\left( s \right) \cdot \mathrm{i}} \right) - \frac{1}{-s + 1}\\ \int_{1}^{\infty} \frac{1}{n^{s}} \operatorname{d}n &= \infty \Rightarrow \text{no convergence}\\ &\Rightarrow \infty < \sum_{{n} = {1}}^{\infty} \frac{1}{n^{s}} \Rightarrow \text{no convergence of } \sum_{{n} = {1}}^{\infty} \frac{1}{n^{s}}\\ \end{align*} $$ for $\Re\left( s \right) = 1$ and $\Im\left( s \right) \ne 0$: $$ \begin{align*} F\left( n \right) = \int_{0}^{x} n^{-1 - \Im\left( s \right) \cdot \mathrm{i}} \operatorname{d}n &\overbrace{=}^{\text{power rule}} \frac{1}{-\Im\left( s \right) \cdot \mathrm{i}} \cdot n^{-\Im\left( s \right) \cdot \mathrm{i}}\\ F\left( n \right) = \int_{0}^{x} n^{-1 - \Im\left( s \right) \cdot \mathrm{i}} \operatorname{d}n &= \frac{1}{\Im\left( s \right)} \cdot n^{-\Im\left( s \right) \cdot \mathrm{i}} \cdot \mathrm{i}\\ \int_{1}^{\infty} \frac{1}{n^{s}} \operatorname{d}n &= \lim_{{x} \to {\infty}} \left( \frac{1}{\Im\left( s \right)} \cdot x^{-\Im\left( s \right) \cdot \mathrm{i}} \cdot \mathrm{i} \right) - \frac{1}{\Im\left( s \right)} \cdot 1^{-\Im\left( s \right) \cdot \mathrm{i}} \cdot \mathrm{i}\\ \int_{1}^{\infty} \frac{1}{n^{s}} \operatorname{d}n &= \frac{1}{\Im\left( s \right)} \cdot \lim_{{x} \to {\infty}} \left( x^{-\Im\left( s \right) \cdot \mathrm{i}} \right) \cdot \mathrm{i} - \frac{1}{\Im\left( s \right)} \cdot \mathrm{i}\\ \int_{1}^{\infty} \frac{1}{n^{s}} \operatorname{d}n &= \frac{1}{\Im\left( s \right)} \cdot \left( \lim_{{x} \to {\infty}} \left( x^{-\Im\left( s \right) \cdot \mathrm{i}} \right) - 1 \right) \cdot \mathrm{i}\\ \int_{1}^{\infty} \frac{1}{n^{s}} \operatorname{d}n &= \frac{1}{\Im\left( s \right)} \cdot \left( \lim_{{x} \to {\infty}} \left( x \right)^{-\Im\left( s \right) \cdot \mathrm{i}} - 1 \right) \cdot \mathrm{i}\\ \int_{1}^{\infty} \frac{1}{n^{s}} \operatorname{d}n &= \frac{1}{\Im\left( s \right)} \cdot \left( \lim_{{x} \to {\infty}} \left( x \right)^{-\left| \Im\left( s \right) \right| \cdot \mathrm{i}} - 1 \right) \cdot \mathrm{i} \tag{if Im(s) > 0}\\ \int_{1}^{\infty} \frac{1}{n^{s}} \operatorname{d}n &= \frac{1}{\Im\left( s \right)} \cdot \left( \lim_{{x} \to {\infty}} \left( \frac{1}{x} \right)^{\left| \Im\left( s \right) \right| \cdot \mathrm{i}} - 1 \right) \cdot \mathrm{i} \tag{if Im(s) > 0}\\ \int_{1}^{\infty} \frac{1}{n^{s}} \operatorname{d}n &= \frac{1}{\Im\left( s \right)} \cdot \left( \underbrace{0^{\left| \Im\left( s \right) \right| \cdot \mathrm{i}}}_{0^{\mathrm{i}} \text{ is not defined}} - 1 \right) \cdot \mathrm{i} \Rightarrow \text{indeterminate} \tag{if Im(s) > 0}\\ &\Rightarrow \text{no convergence of } \sum_{{n} = {1}}^{\infty} \frac{1}{n^{s}} \tag{if Im(s) > 0}\\ \\ \int_{1}^{\infty} \frac{1}{n^{s}} \operatorname{d}n &= \frac{1}{\Im\left( s \right)} \cdot \left( \lim_{{x} \to {\infty}} x^{\left| \Im\left( s \right) \right| \cdot \mathrm{i}} - 1 \right) \cdot \mathrm{i} \tag{if Im(s) < 0}\\ \int_{1}^{\infty} \frac{1}{n^{s}} \operatorname{d}n &= \frac{1}{\Im\left( s \right)} \cdot \left( \lim_{{x} \to {\infty}} e^{\ln\left( x \right) \cdot \left| \Im\left( s \right) \right| \cdot \mathrm{i}} - 1 \right) \cdot \mathrm{i} \tag{if Im(s) < 0}\\ \int_{1}^{\infty} \frac{1}{n^{s}} \operatorname{d}n &= \frac{1}{\Im\left( s \right)} \cdot \left( \underbrace{\lim_{{x} \to {\infty}} \cos\left( \ln\left( x \right) \cdot \left| \Im\left( s \right) \right| \right)}_{\lim_{{u} \to {\infty}} \cos\left( u \right) \text{ is not defined}} + \underbrace{\lim_{{x} \to {\infty}} \sin\left( \ln\left( x \right) \left| \cdot \Im\left( s \right) \right| \right)}_{\lim_{{u} \to {\infty}} \sin\left( u \right) \text{ is not defined}} \cdot \mathrm{i} - 1 \right) \cdot \mathrm{i} \Rightarrow \text{indeterminate} \tag{if Im(s) < 0}\\ &\Rightarrow \text{no convergence of } \sum_{{n} = {1}}^{\infty} \frac{1}{n^{s}} \tag{if Im(s) < 0}\\ \end{align*} $$

But you can also consider these cases as special cases p-series $\left( \sum_{{n} = {1}}^{\infty} \frac{1}{n^{p}}, p > 0 \right)$. This could save you a lot of work for non-complex $s$.

The Art Of Repetition
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Check that $$\lim_{N\to \infty}\sum_{n=1}^N (n^{-s} - \int_n^{n+1} x^{-s}dx)$$ converges for $\Re(s) >0$.

So $\sum_{n=1}^\infty n^{-s}$ converges iff $\int_1^\infty x^{-s}dx$ converges.

reuns
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It is divergent for $\Re(s)\leq 0$ by $n$-the term test, so we consider only $0<\Re(s)\leq 1$. It is divergent for $s=1$.

Let us take $a_k=1$ and $b_k=\frac{1}{k^s}$ in the Abel summation formula, $$\sum_{k=1}^na_kb_k=A_nb_{n+1}-\sum_{k=1}^n A_k\left(b_{k+1}-b_k\right)$$ where $A_n=\sum_{k=1}^na_k$. Then we have $$\sum_{k=1}^n\frac{1}{k^s}=\frac{n}{(n+1)^s}-\sum_{k=1}^n k\left(\frac{1}{(k+1)^s}-\frac{1}{k^s}\right)\tag 1$$ Let $0<\Re(s)<1$ and assume that $\sum_{k=1}^n\frac{1}{k^s}$ is convergents then since the first term of $(1)$ on the RHS is divergent the series on the RHS must be divergent. But, it has the same character of $\sum_{k=1}^n\frac{1}{k^s}$, because $$\frac{1}{(k+1)^s}-\frac{1}{k^s}=k\left(k^{-s}+{-s\choose 1}k^{-s-1}+...-k^{-s}\right)=-sk^{-s}+...$$ Hence, we have a contradiction, I suppose. At least I tried. All remains is $\Re(s)=1$ and $s\neq 1$. This link gives an idea for $s=1+i$.

Bob Dobbs
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