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I want to show that if $a_{n}\downarrow 0$ then $\displaystyle \sum_{n=1}^{\infty}\frac{a_{n}}{n}<+\infty \Leftrightarrow \sum_{n=1}^{\infty}\Delta a_{n}\log n<+\infty.$

I am stuck to prove this result. Please help me out.

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    What is $\Delta a_n\log n$?2012-06-03
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    $\Delta a_{n} \log n= (a_{n}-a_{n+1}) \log n$2012-06-03
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    $\displaystyle \sum_{k=m}^{n}a_{k}b_{k}=\sum_{k=m}^{n-1}\Delta a_{k}B_{k}+a_{n}B_{n}-a_{m}.B_{m-1}$, where $B_{n}=\sum_{k=0}^{n}b_{k}$.So i have to use this formula but how?2012-06-03
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    Take $b_k=1/k$, then $B_n=\sum\limits_{k=1}^n 1/k\approx \log n$2012-06-03
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    $$\displaystyle \sum_{k=1}^{n}\frac{a_{k}}{k}=a_{n}B_{n}+\sum_{k=1}^{N-1}\Delta a_{k}B_{k},\tag{1}$$ where $\displaystyle \log n< B_{n}=\sum_{k=1}^{n}\frac{1}{k}<1+\log n$ Let $$\sum_{n=1}^{\infty}\frac{a_{n}}{n}<+\infty.$$ From $(1)$, we have, $$|a_{n}B_{n}+\sum_{k=1}^{n-1}\Delta a_{k}B_{k}|\leq M$$ Which further implies that $$|\sum_{k=1}^{N-1}\Delta a_{k}B_{k}|\leq M$$ $$\therefore \sum_{k=1}^{\infty}\Delta a_{k}B_{k}<+\infty.$$ $$\sum_{k=1}^{\infty}\Delta a_{n}\log n<\sum_{k=1}^{\infty}\Delta a_{n}B_{n}<+\infty.$$ I am proving only one part. Am i right?2012-06-03
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    Yes, you are right2012-06-03
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    @Norbert Can we prove this result without using Abel summation formula?2012-06-03
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    I don't know, Abel transform is not a sledgehammer.2012-06-03

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For every $N\geqslant1$, $$ \sum\limits_{n=1}^N\frac{a_n}n=\sum\limits_{n=1}^{+\infty}\Delta a_n\cdot H_{\min\{n,N\}}\quad\text{with}\quad H_n=\sum\limits_{k=1}^n\frac1k\sim\log n, $$ hence $$ \sum\limits_{n=1}^{N}\Delta a_n\cdot H_n\leqslant\sum\limits_{n=1}^N\frac{a_n}n\leqslant\sum\limits_{n=1}^{+\infty}\Delta a_n\cdot H_n. $$ Using the bounds $H_n\geqslant\log n$ for $n\geqslant1$ and $H_n\leqslant2\log n$ for $n\geqslant2$, one gets $$ \sum\limits_{n=1}^{N}\Delta a_n\cdot \log n\leqslant\sum\limits_{n=1}^N\frac{a_n}n\leqslant\Delta a_1+2\sum\limits_{n=2}^{+\infty}\Delta a_n\cdot \log n. $$