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.
Best Answer
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).