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Let $s_n$ be a sequence of positive real numbers. Clearly if $\sum\frac{1}{s_n\log s_n}$ diverges then $\sum\frac{1}{s_n}$ diverges but is this implication strict?

I.e. is there a sequence such that the first sum converges but the second sum diverges?

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My roommate just answered this! Let $s_n=n(\log n)^p$ with $0<p\leq 1$. Then the second sum diverges (it is a standard case), and for the first sum we have:

$$s_n \log s_n = n (\log n)^p \log(n(\log(n)^p)=n(\log n)^{p+1} +n(\log n)^p \log(\log n)^p$$

So the first sum converges because $\frac{1}{s_n\log s_n}\leq \frac{1}{n(\log n)^{p+1}}$ and $\sum\frac{1}{n(\log n)^{p+1}}$ converges.

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