# Prove that $\sum_{n}\frac{a_n}{n^\beta}$ converges given convergence of $\sum_{n}\frac{a_n}{n^\alpha}$ for $\alpha<\beta$

If the series $$\sum\limits_{n=1}^{\infty}\frac{a_n}{n^\alpha}$$converges, for any $$\beta>\alpha$$, prove that $$\sum\limits_{n=1}^{\infty}\frac{a_n}{n^\beta}$$ also converges.

I suppose that it can be proved with Cauchy convergence theorem, but I failed. Also, I have no idea how to deal with For any $$\beta>\alpha$$,should I compare them with zero? Any help will be appreciated.

• Is nothing known about $a_n$ ? It would be nicer if the $a_i$ were positive. – Gabriel Romon Apr 2 '15 at 11:00
• @LeGrandDODOM It's not sure whether $a_i$ is positive. – Ferry Tau Apr 2 '15 at 11:02

Rewrite $\displaystyle\sum\limits_{n\geq1}\frac{a_n}{n^\beta}=\sum\limits_{n\geq1}\frac{a_n}{n^\alpha}\times \frac{1}{n^{\beta-\alpha}}$
The partial sums of $\displaystyle\sum\limits_{n\geq1}\frac{a_n}{n^\alpha}$ are bounded by the initial assumption and $\displaystyle\frac{1}{n^{\beta-\alpha}}$ is a decreasing sequence that goes to $0$.