Calculating $\lim_{n\to\infty}\sum_{k=1}^{\infty}\frac{(-1)^{k}\arctan(n^{2}k)}{n^{\alpha}+k^{3/2}}$ I am studying for my exam in real analysis and I am having difficulties
with some of the material, I know that the following should be solved
by using the counting measure and LDCT, but I don't know how.

For $\alpha>0,$ Calculate
  $$\lim_{n\to\infty}\sum_{k=1}^{\infty}\frac{(-1)^{k}\arctan(n^{2}k)}{n^{\alpha}+k^{3/2}}$$

I would greatly appreciate it if in the answers you can include all the details
about the theorems used (since there is a high importance for the
arguments for why we can do what we do in each step, and I don't understand
the material good enough to be able to understand that some step is
actually not trivial and uses some theorem)
 A: Note that for terms of serie $\sum_{k=1}^n \frac{(-1)^{k}\arctan(n^{2}k)}{n^{\alpha}+k^{3/2}}$ we have
\begin{align}
\left|\frac{(-1)^{k}\arctan(n^{2}k)}{n^{\alpha}+k^{3/2}}\right|
= 
&
\left| \frac{\arctan(n^{2}k)}{n^{\alpha}+k^{3/2}}\right|
\\
\leq
&
\left| \frac{2\pi}{n^{\alpha}+k^{3/2}}\right|
\\
=
&
\left| \frac{\pi}{\frac{n^{\alpha}+k^{3/2}}{2}}\right|
\\
\leq
&
\frac{\pi}{\sqrt[2\,]{n^{\alpha}\cdot k^{3/2}}}
\\
=
&
\frac{\pi}{n^{\frac{\alpha}{2}}\cdot k^{3/4}}
\\
=
&
\frac{1}{n^{\frac{\alpha}{2}}}\frac{\pi}{k^{3/4}}
\end{align}
Then 
$$
0\leq \left| \sum_{k=1}^n \frac{(-1)^{k}\arctan(n^{2}k)}{n^{\alpha}+k^{3/2}}\right|\leq \sum_{k=1}^n\left|\frac{1}{n^{\frac{\alpha}{2}}}\frac{\pi}{k^{3/4}}\right|
$$
By Squeeze Theorem for Sequences, $\sum_{k=1}^n\left|\frac{1}{n^{\frac{\alpha}{2}}}\frac{\pi}{k^{3/4}}\right|\to 0$ (for all $\alpha>0$) implies $\left|\sum_{k=1}^n \frac{(-1)^{k}\arctan(n^{2}k)}{n^{\alpha}+k^{3/2}}\right| \to 0$
A: Let $a(n,k):=(-1)^k\frac{\arctan(n^2k)}{n^{\alpha}+k^{3/2}}$; then $|a(n,k)|\leqslant \frac{\pi}{2k^{3/2}}$ for each $n$, hence 
$$\left|\sum_{k=1}^{+\infty}a(n,k)\right|\leqslant \sum_{k=1}^N|a(n,k)|+\frac{\pi}2\sum_{k\geqslant N+1}k^{-3/2},$$
which gives that for each integer $N$, 
$$\limsup_{n\to +\infty}\left|\sum_{k=1}^{+\infty}a(n,k)\right|\leqslant\sum_{k\geqslant N+1}k^{-3/2}.$$
Conclude (we actually used dominated convergence theorem).
