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.
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.
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. $