Comparison test for series convergence I have these two series and I've been trying to figure out correct series to compare them to, to prove their convergence.
The first one:$$\sum_{n=0}^{\infty} (\cos n \pi) \left( \sqrt[3]{\sqrt{n}+1}-\sqrt[3]{\sqrt{n}-2}\right)$$
I'm gonna check if it converges absolutely:
$$\sum_{n=0}^{\infty} \left|  \left(\sqrt[3]{\sqrt{n}+1}-\sqrt[3]{\sqrt{n}-2}\right) \left( \frac{(\sqrt{n}+1)^{2/3} + (\sqrt{n}+1)\sqrt{n}-2) + (\sqrt{n}-2)^{2/3}}{(\sqrt{n}+1)^{2/3} + (\sqrt{n}+1)\sqrt{n}-2) + (\sqrt{n}-2)^{2/3}}\right)\right| =$$
$$=\sum_{n=0}^{\infty} \left| \frac{\sqrt{n}+1 -\sqrt{n}+2}{(\sqrt{n}+1)^{2/3} + (\sqrt{n}+1)(\sqrt{n}-2) + (\sqrt{n}-2)^{2/3}} \right| =$$
$$= 3\sum_{n=0}^{\infty} \left| \frac{1}{2(\sqrt{n}-2)^{2/3} +(\sqrt{n}+1)(\sqrt{n}-2)}\right|$$
The second one:$$\sum_{n=1}^{\infty} n^{2-\frac{3}{n}} \frac{\arctan(-3+\pi n)}{\sqrt[3]{\sin 4n + n^9+4n^8+6}}=\sum_{n=1}^{\infty} \frac{n^2 \arctan(-3+\pi n)}{n^{\frac3{n}}\sqrt[3]{\sin 4n + n^9+4n^8+6}}$$
I don't really know how to simplify anything here...
 A: Limited expansions are the way to go (and I am afraid that avoiding them would result in horrible computations...). To study the first series, one can start with the fact that, when $u\to0$,
$$
\sqrt[3]{1+u}=1+au+bu^2+O(u^3),
$$
for some specific constants $a$ and $b$ whose values are unimportant. Hence, using $u=1/\sqrt{n}$, one sees that
$$
\sqrt[3]{\sqrt{n}+1}-\sqrt[3]{\sqrt{n}-2}=\sqrt[6]{n}\left(3au-3bu^2\right)+x_n,
$$
where $x_n=O\left(\sqrt[6]{n}\cdot u^3\right)=O\left(n^{-4/3}\right)$, hence the series
$$
\sum_{n\geqslant1}(-1)^nx_n
$$
converges (absolutely). Now, $\cos(n\pi)=(-1)^n$ and the alternated series theorem shows that the series
$$
\sum_{n\geqslant1}(-1)^nn^{-\alpha}
$$
converges for $\alpha=\frac12-\frac16=\frac13$ and $\alpha=1-\frac16=\frac56$. Finally, the series
$$
\sum_{n\geqslant1}\cos(n\pi)\left(\sqrt[3]{\sqrt{n}+1}-\sqrt[3]{\sqrt{n}-2}\right),
$$
is also
$$
\sum_{n\geqslant1}(-1)^n(3an^{-1/3}-3bn^{-5/6}+x_n),
$$
which converges (but not absolutely).
To study the second series, I suggest to show that the $n$th term is equivalent to
$$
\frac{\pi}{2n}.
$$
