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?
 A: 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.
A: I am not sure about Leibitz, but the terms are decreasing in absolute value, and the series is alternating, so you are good.
A: The convergence of $$\sum_{n=1}^N (-1)^n \dfrac{n^2}{\sqrt{n^5+1}}$$ can be concluded based on alternating test. The general result is termed as generalized alternating test or Dirichlet test and is based on Abel partial summation. We will prove the generalized statement, though this is a bit of a overkill for this problem it is as easy to prove as the alternating test.
Consider the sum $S_N = \displaystyle \sum_{n=1}^N a(n)b(n)$. Let $A(n) = \displaystyle \sum_{n=1}^N a(n)$. If $b(n) \downarrow 0$ and $A(n)$ is bounded, then the series $\displaystyle \sum_{n=1}^{\infty} a(n)b(n)$ converges.
First note that from Abel summation, we have that
$$\sum_{n=1}^N a(n) b(n) = \sum_{n=1}^N b(n)(A(n)-A(n-1)) = \sum_{n=1}^{N} b(n) A(n) - \sum_{n=1}^N b(n)A(n-1)\\
= \sum_{n=1}^{N} b(n) A(n) - \sum_{n=0}^{N-1} b(n+1)A(n) = b(N) A(N) - b(1)A(0) + \sum_{n=1}^{N-1} A(n) (b(n)-b(n+1))$$
Now if $A(n)$ is bounded i.e. $\vert A(n) \vert \leq M$ and $b(n)$ is decreasing, then we have that
$$\sum_{n=1}^{N-1} \left \vert A(n) \right \vert (b(n)-b(n+1)) \leq \sum_{n=1}^{N-1} M (b(n)-b(n+1))\\ = M (b(1) - b(N)) \leq Mb(1)$$
Hence, we have that $\displaystyle \sum_{n=1}^{N-1} \left \vert A(n) \right \vert (b(n)-b(n+1))$ converges and hence $$\displaystyle \sum_{n=1}^{N-1} A(n)  (b(n)-b(n+1))$$ converges absolutely. Now since
$$\sum_{n=1}^N a(n) b(n) = b(N) A(N) + \sum_{n=1}^{N-1} A(n) (b(n)-b(n+1))$$
we have that $\displaystyle \sum_{n=1}^N a(n)b(n)$ converges.
In your case, $a(n) = (-1)^n$. Hence, $$A(N) = \displaystyle \sum_{n=1}^N (-1)^n$$which is clearly bounded.
Also, $b(n) = \dfrac{n^2}{\sqrt{n^5+1}}$ is a monotone decreasing sequence converging to $0$.
Hence, we have that $$\sum_{n=1}^N (-1)^n \dfrac{n^2}{\sqrt{n^5+1}}$$ converges.
