TITLE: Convergence of a serie by comparison criteria
QUESTION [0 upvotes]: I have to look for the convergence of the following series:
$$ \sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n\sqrt{n+1}} $$
I know I have to make use of the comparison criteria but I do not know how to follow from there.
I also see that the numerator would make the series divergent

REPLY [1 votes]: The series is absolutely convergent hence is convergent. The absolute convergent comes from the $\displaystyle \sum_{n=1}^\infty |a_n|$ converges with $a_n = \dfrac{(-1)^n}{n\sqrt{n+1}}$. This is true as you compare $|a_n| \leq n^{-\frac{3}{2}}$.

REPLY [0 votes]: The series converges absolutely because $\sum 1/n^\alpha$ converges for $\alpha > 1$.  Baby Rudin 3.28.  You have a slight variant on $\sum 1/n^{3/2}$.
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