Find out whether series diverges or converges I have this math problem: "determine whether the series converges absolutely, converges conditionally, or diverges."
I can use any method I'd like. This is the series:$$\sum_{n=1}^{\infty}(-1)^n\frac{1}{n\sqrt{n+10}}$$
I though about using a comparison test. But I'm not sure what series I can compare $\sum_{n=1}^{\infty}\frac{1}{n\sqrt{n+10}}$ to.
 A: Converges absolutely
$|u_n|=\frac{1}{n\sqrt{(n+10)}}<\frac{1}{n}.\frac{1}{n^{\frac{1}{2}}}=\frac{1}{n^{\frac{3}{2}}}$
A: We can use Leibniz theorem: if a_n is a monotone sequence that converges to 0, then the series $\sum_{n=1}^{\infty}(-1)^n a_n$ converges. Let´s have $a_n=\dfrac{1}{n\sqrt{n+10}}$. 
So we have to prove that $lim$ an=0 and that it is monotone. If you can show that, you have that it converges   
A: Note that the summand is alternating, and converges monotonically to zero in absolute value. This implies that the series is convergent.
The series, $S = \sum_{n=1}^{\infty}\frac{(-1)^{n}}{n\sqrt{n+10}}$, is absolutely convergent:
$$|S| =|\sum_{n=1}^{\infty}\frac{(-1)^{n}}{n\sqrt{n+10}}| \leq \sum_{n=1}^{\infty} |\frac{1}{n\sqrt{n+10}}| \leq \sum_{n=1}^{\infty}\frac{1}{n\sqrt{n}} = \sum_{n=1}^{\infty}n^{-\frac{3}{2}}$$ 
Since $f(t) := t^{-\frac{3}{2}}$ is positive and decreasing on the set $[1,\infty) \subset \mathbb{R}$, it follows that $\sum_{n=1}^{\infty}f(n)$ converges if and only if $\int_{1}^{\infty}f(t)dt$ converges. Since $\int_{1}^{\infty}f(t)dt = \int_{1}^{\infty}t^{-\frac{3}{2}} = 2$, it follows that the series $\sum_{n=1}^{\infty}n^{-\frac{3}{2}}$ is convergent. Therefore $S$ is absolutely convergent. $\square$ 
