Divergence of the Dirichlet series for the Riemann zeta function for Re s = 1, Im s <> 0 Consider the Dirichlet series $\sum_{n=1}^{\infty} n^{-s}$, with $s=\sigma+it$, $\sigma$ and $t$ real. How can one prove that this series diverges for $\sigma=1$ and $t\neq 0$?
In all the other combinations of values of $\sigma$ and $t$, it is rather easy to determine (and prove) whether the series converges or diverges:


*

*In the case $\sigma=1$ and $t=0$, the Dirichlet series becomes the harmonic series, which, of course, diverges.

*In the case $\sigma>1$ (and any $t$), it is trivial to prove, by the integral test, that the series converges absolutely. 

*The case $\sigma<1$ (and any $t$) is only slightly more complicated than the previous ones. In this case it is to be proven that the series diverges. To this end one uses a version of the Euler-Maclaurin formula, namely
\begin{equation}
\sum_{n=1}^{N}f(n) - f(1) = \int_{1}^{N}f(x)dx+\int_{1}^{N}f'(x)(x-\lfloor x \rfloor)dx,
\end{equation}
where $\lfloor x \rfloor$ is the floor function (returning the greatest integer less than or equal to $x$). By playing with the triangle inequality one gets 
\begin{equation}
\left| \sum_{n=1}^{N}f(n) \right| \geq \left| \int_{1}^{N}f(x)dx\right| - \left| \int_{1}^{N}f'(x)(x-\lfloor x \rfloor)dx \right| - \left|f(1)\right|;
\end{equation}
Now notice that $\left| \int_{1}^{N}f'(x)(x-\lfloor x \rfloor)dx \right|\leq \int_{1}^{N}\left|f'(x)\right|dx$, and that for the relevant function $f$ (namely, $f(x)=x^{-s}$), the latter converges as $N\to +\infty$. However, in the same limit, $\left| \int_{1}^{N}f(x)dx\right| $ diverges to infinity; from the formula above, it then follows that the Dirichlet series itself diverges to infinity.
The difficulty when $\sigma=1$, $t\neq 0$ is that $\left| \int_{1}^{N}f(x)dx\right| $ does not diverge to infinity; instead, we have $\int_{1}^{N}1/x^{1+it}dx=\frac{i}{t}\left(-1+e^{-it\ln N}\right)$, which---in the $N\to \infty$ limit---has an oscillatory divergence. Therefore I suppose the series $\sum_{n=1}^{\infty} n^{-s}$ should also diverge in an oscillatory manner; however, I don't know how to use the second integral, $\int_{1}^{N}f'(x)(x-\lfloor x \rfloor)dx$, in order to prove this.
Perhaps the proof should not use Euler-Maclaurin at all; I just don't know.
 A: I wanted to correct a couple of the inequalities in David Speyer's answer and comments.
I believe one in the answer should read $\left|  \frac{1}{n^s} - \int_{x=n}^{n+1} \frac{dx}{x^s} \right| \leq |s|/(2n^{{Re}(s)+1})$.  (This has now been corrected by Mr. Speyer, omitting the "2" in the denominator, which is fine.)
Also, I believe the one in the following comment should read $|f'(x) (x - \lfloor x \rfloor)| \leq |s|/|x|^{{Re}(s)+1}$.
A: Write 
$$\sum_{n=1}^{N} \frac{1}{n^s} = \sum_{n=1}^{N} \left( \frac{1}{n^s} - \int_{x=n}^{n+1} \frac{dx}{x^s} \right) + \int_{x=1}^{N+1} \frac{dx}{x^s} \quad (*)$$
You want to know whether the limit of the left hand side of $(*)$ exists, as $N \to \infty$. Now, you can check that
$$\left|  \frac{1}{n^s} - \int_{x=n}^{n+1} \frac{dx}{x^s} \right| \leq \frac{|s|}{n^{\mathrm{Re}(s)+1}}.$$
So 
$$\sum_{n=1}^{\infty}  \left( \frac{1}{n^s} - \int_{x=n}^{n+1} \frac{dx}{x^s} \right)$$
is absolutely convergent and the limit of the first term on the right hand side of $(*)$ exists. So the question reduces to whether or not
$$\lim_{N \to \infty} \int_{x=1}^{N+1} \frac{dx}{x^s}$$
exists. We can compute the integral explicitly, and see that it does not.
