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I feel like the answer is that yes it does converge, however I have not been successful in finding a bound on the series. Additionally, I have not been able to find a series $\sum a_n$ such that it will diverge when you multiply each term by $\log(a_n)$.

I am not sure what else to do because just knowing that the original series converges doesn't let me apply any of the tests such as the ratio or root tests on the series.

Any advice or tips would be appreciated.

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Hint: Try $a_n = \frac{1}{n \log^2 n}$

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  • $\begingroup$ It still works as $\log \frac{1}{n \log^2 n} = - \log n - \log \log^2 n$ $\endgroup$
    – Michael Biro
    Mar 28, 2016 at 4:09
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    $\begingroup$ Ok got it . Nice answer. $\endgroup$
    – RRL
    Mar 28, 2016 at 4:10

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