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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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    You can't, of course, show $\,a_n\log n\leq 0\,$ . What you probably meant is you want to show $\,a_n\log n\leq b_n\,$ , with $\,b_n\xrightarrow [n\to\infty]{} 0\,$ and then use the squeeze theorem2012-07-20
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    Can you please explain briefly!2012-07-20
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    It is *a fact* that $\,a_n\log n\geq 0\,$, period. If you show the opposite inequality then $\,a_n\log n =0\,$ identically, what of course isn't true unless $\,a_n=a_{n+1}\,$ , which is a rather sharp restriction on the sequence $\,\{a_n\}\,$.2012-07-20
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    @DonAntonio for prove $a_{n}\log n\leq b_{n}$. what i have to take $b_{n}$?2012-07-20
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    That I don't know, @Kns...that'd solve your problem. :) Perhaps there's another way, there's need to make some further thinking.2012-07-20
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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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    By the way, this shows that you could use any positive monotonically increasing function instead of $\log n$.2012-07-20
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    How $\displaystyle \sum_{n=n_{0}}^{\infty} \Delta a_{n}=a_{n_{0}}-\lim_{n\to\infty}a_{n}$?2012-07-20
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    @Kns: Take the limit for $m\to\infty$ of $\displaystyle \sum_{n=n_{0}}^m \Delta a_{n}=a_{n_{0}}-a_{m+1}$.2012-07-20
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    @Kns This is an example of a telescoping sum: $$(b_1-b_2)+(b_2-b_3)+ ... + (b_{n-1}-b_n) = b_1-b_n$$2012-07-20
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    +1 Nice!. But shouldn't we write $$\lim_{k\to\infty}a_{n_0+k}$$ inside those parentheses? Of course, the result is the same as this is the limit of a subsequence of a converging-to-zero sequence...2012-07-20
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    $\lim a_{n_0+k} = \lim a_{k}$ @DonAntonio2012-07-20
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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