How can I bound $\left|\sum_{k=1}^n \frac{(-1)^{k+1}}{k^{bi}}\right|$ where $b\gt 0$? Let $b\gt 0$. I tried
$$S=\left|\displaystyle\sum_{k=1}^n \frac{(-1)^{k+1}}{k^{bi}}\right|$$
converges if
$$T=\left|\displaystyle\sum_{k=1}^n \frac{(-1)^{k+1}}{\left|k^{bi}\right|}\right|$$
converges. And
$$T=\left|\displaystyle\sum_{k=1}^n (-1)^{k+1}\right|\le 1.$$
Is this correct?
 A: Let me put a quick answer leaving out details that are standard in ANT:
The sum in the post is trivially bound by $\sqrt b$ up to some logarithmic factors, 
(which is precisely what one expects by the RZ bound on the imaginary axis which is O($\sqrt b$) up to negligible logarithmic factors by the functional equation) and this follows from the second derivative test immediately, as for $f(x)= \frac{x}{2}+ \frac{b \log x}{2 \pi}$ for which the original sum is $-\sum_{k=1}^{n} {e^{2\pi i f(k)}}$, the second derivative is ~ $\frac{b}{x^2}$, so the second derivative test gives the sum ~ $\sqrt b$ ( there are details like splitting it in $\log_{2} n$ binary parts on $n$, s.t $M \le k \le 2M$ etc but they are standard). This immediately by partial summation gives the convergence of $\eta (s)=\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n^s}$ for $\Re s >0$, uniform in any $\Re s > a >0$ 
A bound of $\sqrt n$ is equivalent to the Lindelof conjecture (technically we need the one below to account for the case $b >> e^n$ say):
$\left|\sum_{k=1}^n \frac{(-1)^{k+1}}{k^{bi}}\right| \le C_{\epsilon}n^{\frac{1}{2} + \epsilon}(2+b)^{\epsilon}$, for all $n \ge 1, b>0, \epsilon >0$
The hard part is beating the $\min ({n, \sqrt b})$ bound, especially when $n << \sqrt b$ when we need moderately advanced results from RZ theory (Riemann Siegel say) or from exponent pairs theory (A process) say. 
But for convergence of $\eta$ as noted, we do not need this stuff since that is concerned with high $n$ compared with $b$ and there indeed there are trivial (in context) bounds
(edit later about exponential sum $S = \sum_{M \le k \le 2M}e^{2\pi i f(k)}$ estimates - there are two basic ones and then they can be refined using various methods; in both cases we assume $f$ smooth enough and $f',f''$ monotonic and with $||x||$ the distance from $x$ to closest integer we have;
Thm 1: Kuzmin-Landau: If $||f'(x)||\ \ge \delta >0, x \in [M,2M]$, then $S=O(\frac{1}{\delta})$ with universal constant (I think 4 works)
(Mordell has a very neat proof of this in a very general sequence setting which is in the proceedings of the ICM 1954 and is also presented in the Iwaniec-Kowalski ANT bible if I remember right)
Thm 2: Second derivative test (due to Van der Corput I think); if now $\lambda \le |f''(x)| \le a\lambda, x \in [M,2M], a, \lambda >0$, then $S=O(aM\sqrt{\lambda}+\frac{1}{\sqrt \lambda})$ with again universal constant (I think 8 works here). 
(note also that we need $\lambda << 1$ as otherwise the trivial estimate will do, while the second term is required if $\lambda$ is very small). 
This is proved by carefully splitting $[M,2M]$ in intervals ($aM\lambda +2$ in number) where $||f'|| \ge \delta$ for some small parameter $\delta$, and intervals of length at most $\frac{2\delta}{\lambda}$ ($aM\lambda+1$ in number) applying K-l on first kind, trivial estimates on second and making the two estimates equal with $\delta = \sqrt \lambda$
