If $\;\sum\limits_{n=1}^\infty {a_n}\;$ converges,

does the following also converge? $$\sum_{n=1}^\infty {a_n}{\ln{a_n}}$$

When I first saw this problem, it was easy.

So I tried comparison test and limit comparison test.

But i couldn't solve it.

Could you help me?


First of all, we have to assume that $a_n>0$ for each $n$, otherwise $\log a_n$ wouldn't make sense. The problem is that $a_n$ can go to $0$ fast enough to make the series $\sum_n a_n$ convergent, but the fact that $\log a_n$ goes to $-\infty$ can spoil the convergence.

Indeed, define $a_n:=\frac 1{n(\log n)^2}$. Then $\sum_n a_n$ is convergent. On the other hand,
$$|a_n\log a_n|\sim \frac 1{n\log n},$$ hence the series $\sum_n a_n\log a_n$ is divergent ($a_n\log a_n\leqslant 0$).

  • $\begingroup$ How can you find a couterexample easily? $\endgroup$ – user148928 May 9 '14 at 14:51
  • 2
    $\begingroup$ We want $\sum_n a_n$ to be convergent, but with $a_n\to 0$ not too fast because $a_n\log a_n$ will be too small to diverge. Series of the form $\sum_n n^{-1}(\log n)^{-b}$ converge for $b>1$ and diverge otherwise, hence these series are good candidates. $\endgroup$ – Davide Giraudo May 9 '14 at 15:03
  • $\begingroup$ Nice! Thank you for your help! $\endgroup$ – user148928 May 9 '14 at 15:36

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