The $\sum \limits_{n=0}^{\infty} \dfrac {a_n}{n} $ where $a_n$ is any sequence that meets the condition: $\lim \limits_{n \to \infty} a_n=0$.

I have tried to apply the different tests available such as the limit comparison test and divergence test but I am not sure how to rigorously prove this statement. Any help would be appreciated.

  • 2
    $\begingroup$ The statement is false. $\endgroup$ – user296602 Mar 10 '16 at 5:21
  • $\begingroup$ I heavily edited your question. Please be sure I did not change its meaning. $\endgroup$ – zz20s Mar 10 '16 at 5:23
  • $\begingroup$ The answer you accepted does not answer your edited question. $\endgroup$ – fosho Mar 10 '16 at 6:04
  • 1
    $\begingroup$ The sum should start at $n=1$ since $\frac{a_n}n$ is bad for $n=0$. $\endgroup$ – robjohn Mar 10 '16 at 6:10

This is not an answer to the question, but it shows that the question cannot be answered.

If $$ a_n=\frac1{\log(n+1)} $$ then the series $$ \sum_{n=1}^\infty\frac{a_n}n $$ does not converge.

  • $\begingroup$ Apply the Cauchy Condensation Test: If $ f(n) \geq f(n+1)\geq 0$ for all but finitely many $n$ then $\sum_{n=1}^{\infty} f(n)$ converges iff $\sum_{n=1}^{\infty}[2^n f(2^n)]$ converges. $\endgroup$ – DanielWainfleet Mar 10 '16 at 7:29

This not true, see $a_n = \frac{1}{\sqrt{n}}$.

  • $\begingroup$ Is this supposed to be a counterexample? The series $\sum n^{-3/2}$ converges. $\endgroup$ – RRL Mar 10 '16 at 6:25
  • $\begingroup$ When this answer was posted, OP was asking a completely different question @RRL $\endgroup$ – ASKASK Mar 10 '16 at 7:15
  • $\begingroup$ Got it. Thanks. $\endgroup$ – RRL Mar 10 '16 at 7:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.