# 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...

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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 $$\frac{\pi}{2n}.$$