Convergence of $\sum\limits_{n=0}^{\infty}(-1)^n\frac{n^2}{\sqrt{n^5+1}}$ I want to check, whether $\sum\limits_{n=0}^{\infty}(-1)^n\frac{n^2}{\sqrt{n^5+1}}$ converges or diverges.
I tried to use Leibniz's test :
$|a_n|= \frac{n^2}{\sqrt{n^5+1}} = \frac{n^2}{\sqrt{n^4(n+\frac{1}{n^4})}} = \frac{n^2}{n^2\sqrt{n+\frac{1}{n^4}}} = \frac{1}{\sqrt{n+\frac{1}{n^4}}}$
So $\lim\limits_{n \rightarrow \infty}{{\frac{1}{\sqrt{n+\frac{1}{n^4}}} = 0}} $
$1>|\frac{a_{n+1}}{a_n}|= \frac{(n+1)^2}{\sqrt{(n+1)^5+1}} \frac{\sqrt{n^5+1}}{n^2}= \frac{2n+1 \sqrt{n^5+1}}{\sqrt{(n+1)^5+1}}= \frac{n^2+2n+1 \sqrt{n+\frac {1}{n^4}}}{(n+1)^2\sqrt{n+1+1}} = \frac {\sqrt{n+\frac {1}{n^4}}}{\sqrt{n+2}}$
So $\sum\limits_{n=0}^{\infty}(-1)^n\frac{n^2}{\sqrt{n^5+1}}$ converges.
Could somebody please check my solution?

We have
$$(n^5+1)^{-1/2}=\frac{1}{n^{5/2}}\left(1+\frac{1}{n^5}\right)^{-1/2}=\frac{1}{n^{5/2}}\left(1+O\left(\frac{1}{n^5}\right)\right)$$
hence
$$(-1)^n\frac{n^2}{\sqrt{n^5+1}}=\underbrace{\frac{(-1)^n}{n^{1/2}}}_{=u_n}+\underbrace{O\left(\frac{1}{n^{11/2}}\right)}_{=v_n}$$
the series $\displaystyle\sum_n u_n$ is convergent by the Leibniz theorem and the series $\displaystyle\sum_n v_n$ is also convergent by comparison with the Riemann series. Conclude.
