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I'm aware there are been several similar questions, and some great answers with hints on how to prove this, for example here and here.

But I've never really seen a detailed proof of the absolute convergence of the Riemann zeta function in the half-plane $\Re(s)> 1$.

I hope there's interest on a detailed proof of this for reference. Here is my attempt. Please, fill in any detail that might be missing, or point out any mistake!

On $\Re(s)=\sigma> 1$, we have

$$\sum_{n=1}^{\infty}\bigg|\frac{1}{n^s}\bigg|=\sum_{n=1}^{\infty}\frac{1}{n^\sigma}$$

and

$$\frac{1}{n^\sigma}\leq\frac{1}{n^{1+\epsilon}}$$

for any $\epsilon >0$, so by the direct comparison test, if

$$\sum_{n=1}^{\infty}\frac{1}{n^{1+\epsilon}}$$

converges absolutely, then so do $\sum_{n=1}^{\infty}|1/n^s|$ and $\zeta(s)$.

By the integral test, the convergence of $\sum_{n=1}^{\infty}1/n^{1+\epsilon}$ is equivalente to the finitude of the integral

$$\int_1^\infty \frac{1}{n^{1+\epsilon}}=\bigg[-\frac{1}{\epsilon x^\epsilon}\bigg]=\frac{1}{\epsilon}$$

which holds for any $\epsilon < \infty$.

Any correction or improvement is welcomed! Thanks in advance.

Community
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G. Schiele
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  • The proof seems to be working (although there are some typos in the last integral), but I don't see the need to introduce $\epsilon$. Working directly with $\sigma$ gives exactly the same result. – b00n heT Aug 19 '16 at 19:24
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    As I understand it, Dirichlets test is if you have $\sum_{n}a_nb_n$ then it converges if the partial sums of either $a_n$ or $b_n$ are bounded, and the limit of the other goes to zero and is monotonically decreasing. This is the version that's in Blue Rudin and is proved quite readily from the expression for summation by parts. What is the generalization you're thinking of if not this? – Benjamin Gadoua Aug 20 '16 at 00:23
  • @BenjaminGadoua SEE THIS ANSWER, which details, references, and applies the generalization. That said, the previous comment was not relevant to the application of the generalization. So, I have deleted it accordingly. -Mark – Mark Viola Aug 20 '16 at 04:16
  • for $\zeta(s)$ and more generally Dirichlet series, the most useful is the Abel summation formula (some sort of integration by parts) $\sum_{n=1}^N n^{-s} = s \int_1^N \lfloor x \rfloor x^{-s-1}dx + N^{1-s}$ – reuns Aug 20 '16 at 09:28
  • @benjamingadoua I've posted a solution that uses the Euler-Maclaurin Summation Formula to show explicitly convergence for $\sigma>1$ and diveegence otherwise. – Mark Viola Aug 20 '16 at 15:21

1 Answers1

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Here is another approach. Let $s=\sigma +i \omega$. Then, we can write

$$\begin{align} \sum_{n=1}^\infty \frac{1}{n^s}&=\sum_{n=1}^\infty \frac{e^{-i\omega \log(n)}}{n^\sigma }\\\\ &=\sum_{n=1}^\infty \frac{\cos(\omega \log(n))}{n^\sigma }-i\sum_{n=1}^\infty \frac{\sin(\omega \log(n))}{n^\sigma }\\\\ \end{align}$$

Next, we use the Euler–Maclaurin Summation Formula (EMSF) to write for $\sigma \ne 1$, $\sigma >0$

$$\sum_{n=1}^N \frac{\cos(\omega \log(n))}{n^\sigma }=N^{1-\sigma}\left(\frac{(1-\sigma)\cos(\omega \log(N))}{(1-\sigma)^2+\omega^2}+\frac{\omega \sin(\omega \log(N))}{(1-\sigma)^2+\omega^2}\right)+K_1+O\left(N^{-\sigma}\right)$$

$$\sum_{n=1}^N \frac{\sin(\omega \log(n))}{n^\sigma }=N^{1-\sigma}\left(\frac{(1-\sigma)\sin(\omega \log(N))}{(1-\sigma)^2+\omega^2}-\frac{\omega \cos(\omega \log(N))}{(1-\sigma)^2+\omega^2}\right)+K_2+O\left(N^{-\sigma}\right)$$

for some constants $K_1$ and $K_2$ (that $K_1$ and $K_2$ exist can be shown by analyzing the remainder term in the EMSF). Note, in arriving at the series, we used integration by parts twice in applying the EMSF.

Obviously, we find that for $0<\sigma<1$, the series diverge (for $\omega=0$, only the cosine series diverges), while for $\sigma >1$, the series converge. Finally, it is easy to carry out the analysis for the case $\sigma =1$ to see that the series diverge for $\sigma =1$ (See THIS ANSWER).

Mark Viola
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  • Have you seen the proof of convergence given by Lang in his algebraic number theory for general zeta functions that follows from some Tauberian Theorems? actually seems pretty similar to this. – Benjamin Gadoua Aug 20 '16 at 18:11
  • @BenjaminGadoua Ben, no; I've never heard of the book by Lang and in fact, I've not studied number theory. This is simply a way forward using the EMSF, which is an extremely powerful tool that seems to be underused and seldom taught. -Mark – Mark Viola Aug 20 '16 at 18:49