Prove that $\sum_{n\mathop=1}^\infty \frac 1 {n^{1+i}}$ is divergent I am struggling with this analysis exercise:

Prove that $\displaystyle \sum_{n\mathop=1}^\infty \dfrac 1 {n^{1+i}}$ is divergent but the limit $\displaystyle \lim_{t \to 1^+} \sum_{n\mathop=1}^\infty \frac 1 {n^{t+i}}$ exists.

I know the first sum is not absolutely convergent because $\sum \frac 1 n$ diverges by the integral test. How how do I prove it is divergent? (i.e. not conditionally convergent). Would appreciate hints for the 2nd part too
 A: For $s \in \mathbb{C}\setminus \{1\}$, Abel's sum formula applied to $a_n = 1$ and $f(x) = x^{-s}$ yields
\begin{align}
\sum_{n \leqslant x} \frac{1}{n^s} &= \frac{\lfloor x\rfloor}{x^s} + s\int_1^x \frac{\lfloor t\rfloor}{t^{s+1}}\,dt \\
&= \frac{x - \{x\}}{x^s} + s\int_1^x \frac{t - \{t\}}{t^{s+1}}\,dt \\
&= x^{1-s} + s\int_1^x \frac{dt}{t^s} - \frac{\{x\}}{x^s} - s\int_1^x \frac{\{t\}}{t^{s+1}}\,dt \\
&= x^{1-s} + \frac{s}{s-1}\bigl(1 - x^{1-s}\bigr) - \frac{\{x\}}{x^s} - s\int_1^x \frac{\{t\}}{t^{s+1}}\,dt \\
&= \frac{x^{1-s}}{1-s} + \frac{s}{s-1} - \frac{\{x\}}{x^s} - s\int_1^x \frac{\{t\}}{t^{s+1}}\,dt\,.
\end{align}
When $\operatorname{Re} s > 0$, we have $\{x\}\cdot x^{-s} \to 0$, and
$$\lim_{x\to \infty} \int_1^x \frac{\{t\}}{t^{s+1}}\,dt$$
exists, so the convergence of the series depends on whether $x^{1-s}$ converges. Clearly that is the case only for $\operatorname{Re} s > 1$. For $\operatorname{Re} s = 1$, the partial sums approach a limiting circle with radius $1/\operatorname{Im} s$. Hence the series
$$\sum_{n = 1}^{\infty} \frac{1}{n^{1+i}}$$
diverges.
For $\operatorname{Re} s > 1$, the above computation yields
$$\zeta(s) = \frac{s}{s-1} - s\int_1^{\infty} \frac{\{t\}}{t^{s+1}}\,dt\,,$$
and the dominated convergence theorem shows the existence of
$$\lim_{t \to 1^+} \zeta(t+i) = 1 - i - (1+i)\int_1^{\infty} \frac{\{t\}}{t^{2+i}}\,dt\,.$$
A: Note
$$\frac{1}{n^{1+i}} = \frac{\cos (\ln n) - i\sin (\ln n)}{n}.$$
Concentrating on the real part, we have for $m\in \mathbb N$
$$\tag 1 \sum_{e^{2\pi m}< n < e^{2\pi m+\pi/4}}\frac{\cos (\ln n)}{n} >\frac{1}{\sqrt 2}\sum_{e^{2\pi m}< n < e^{2\pi m+\pi/4}}\frac{1}{n}.$$
The last sum is approximately
$$\int_{e^{2\pi m}}^{e^{2\pi m}+\pi/4} \frac{dx}{x}=\frac{\pi}{4}.$$
If the series $\sum 1/n^{1+i}$ converged, it's real part would converge, hence by the Cauchy crierion, the sums in $(1)$ would converge to $0$ as $m\to \infty.$ Since they don't,$\sum 1/n^{1+i}$ diverges.
Of course the "approximately" above needs to be examined, but all is well  and I'll leave it to you to make sure that's true.
