Does $\sum_{k=1}^{\infty} k^{1+\frac{1}{k}}\frac{\arctan{(3+\pi k)}}{\sqrt{k^4+2k-\sin{(k)}}}$ converge? Does $\sum_{k=1}^{\infty} k^{1+\frac{1}{k}}\frac{\arctan{(3+\pi k)}}{\sqrt{k^4+2k-\sin{(k)}}}$ converge?
I can't come up with any good methods to solve this.
Both D'Alembert's and Cauchy's criterions answer is 1 so it can't be solved in this way. It might be solved using Abel's theorem but I can't find any solutions. 

Define, for every $k \in \mathbb{N}$, $a_k := k^{1+\frac{1}{k}}\frac{\arctan(3+\pi k)}{\sqrt{k^4+2k-\sin(k)}}$ and observe that, as $k\to+\infty$, we have
$$
\begin{align}
\arctan(3+\pi k ) &\sim \frac{\pi}{2}, \\
\sqrt{k^4+2k-\sin(k)} &\sim k^2, \\
 k^{1+1/k} = k  e^{1/k \log (k) } &\sim k.
\end{align}
$$
Therefore $a_k \sim \frac{\pi}{2} \frac{k}{k^2} = \frac{\pi}{2} \frac{1}{k}$. Since $a_k \geq 0$ for every $k \in \mathbb{N}$ and since the series $\sum_1^\infty \frac{1}{k}$ diverges, we conclude that $\sum_1^\infty a_k$ diverges by the Limit Comparison Test.
