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.

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

Apply Abel Discriminance to this problem.

  • $\begingroup$ Yes,I get it.Thanks. $\endgroup$ – Ferry Tau Apr 2 '15 at 11:08
  • 4
    $\begingroup$ What is the Abel Discriminance? I've never heard of this before and a google search yields nothing except a paper behind a paywall. $\endgroup$ – JessicaK Apr 2 '15 at 11:08
  • $\begingroup$ Oh, I mean "Abel test" link $\endgroup$ – Jay Apr 2 '15 at 11:16

Dirichlet test actually does the trick.

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 $0$.


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