How to prove that $\sum_{n=1}^{\infty} \frac{n^{\alpha}}{n(n+1)}$ converges for $\alpha < 1$? I am trying to prove that $\sum_{n=1}^{\infty} \frac{n^{\alpha}}{n(n+1)} < \infty \iff \alpha < 1$, for $\alpha$ a real number. 
Surely if $\alpha \geq 1$, then $\frac{n^{\alpha}}{n(n+1)} \geq \frac{1}{n+1} \Rightarrow \sum_{n=1}^{\infty} \frac{n^{\alpha}}{n(n+1)} = \infty$. However, I'm not sure how to prove convergence if $\alpha < 1$. 
Thanks!
 A: If the series is convergent, then notice that 
$$ \frac{1}{n^{1- \alpha}(n+1)} = \frac{1}{n^{2-\alpha}+n^{1-\alpha}}$$
Since this converges, we know by the p-series that $2-\alpha>1$ which means $1 > \alpha $
A: You basically just turn your argument on its head. You have $\sum \frac{1}{n^{1-\alpha}(1+n)} \leq \sum \frac{1}{n^{2 - \alpha}}$, and your assumption that $\alpha < 1$ implies $2- \alpha > 1$, so life is good, by $p$-series now.
You could also use a limit comparison argument, if you like, using the same idea.
A: The series 
$$
\sum_{n \geq 0}\frac{1}{n^{2-\alpha}}
$$
converges, because it is a $p$-series with $p = 2 - \alpha   > 2-1 = 1$. Now, by the limit test, 
$$
\lim_{n \to \infty} \frac{n^{2-\alpha}n^{\alpha}}{n(n+1)} = \lim_{n \to \infty} \frac{n^2}{n^2(1+\frac{1}{n})} = \lim_{n \to \infty} \frac{1}{1+\frac{1}{n}} = 1 
$$
and so $\sum_{n \geq 0 }\frac{n^\alpha}{n(n+1)} < \infty$.
A: As an alternative answer - 
If $\alpha > 1$, then for $n$ an integer, $n^\alpha > n \Rightarrow \frac{n^{\alpha}}{n} > 1$. Then $\frac{n^{\alpha}}{n(n+1)} > \frac{1}{n+1}$, and so $\sum_{n=1}^{\infty} \frac{n^{\alpha}}{n(n+1)}$ diverges since $\sum_{n=1}^{\infty} \frac{1}{n+1}$ diverges, by the comparison test.
