Calculating $\lim_{n\to\infty}\sum_{k=1}^{\infty}\frac{(-1)^{k}\arctan(n^{2}k)}{n^{\alpha}+k^{3/2}}$

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 $\left|\sum_{k=1}^n \frac{(-1)^{k}\arctan(n^{2}k)}{n^{\alpha}+k^{3/2}}\right| \to 0$

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).