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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.

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    $\begingroup$ 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
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    $\begingroup$ 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
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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.

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

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  • $\begingroup$ 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

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