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Suppose i have $a_{n}\downarrow 0$ and $\displaystyle \sum_{n=1}^{\infty}\Delta a_{n}\log n<+\infty$, where $\Delta a_{k}=a_{k}-a_{k+1}$ and $a_{n}\downarrow 0$ means $a_{n}$ is decreasing and convergent to $0$.

I want to show that $a_{n}\log n \to 0$ as $n\to\infty.$ Clearly, $a_{n}\log n\geq 0$. I need to show that $a_{n}\log n\leq 0$ but i don't know how to solve this!

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    Can we take $\displaystyle b_{n}=\sum_{k=1}^{n} \frac{1}{k}$, where \log n< b_{n}<1+\log n?2012-07-20

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$\sum_{n=n_0}^\infty\Delta a_n\log n\ge\sum_{n=n_0}^\infty\Delta a_n\log n_0=\log n_0\sum_{n=n_0}^\infty\Delta a_n=\left(a_{n_0}-\lim_{n\to\infty}a_n\right)\log n_0=a_{n_0}\log n_0$

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    I know that and *precisely* that is what I wrote in my past comment, @ThomasAndrews, thanks. My point was only for the sake of the *very* accurate notation of the running index, that's all.2012-07-20