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Suppose $a_n>0$ for all $n\in \mathbb{N}$, and consider two series $$\sum_{n=1}^{\infty}\frac{1}{a_n},\qquad \sum_{n=1}^{\infty}\frac{n}{a_1+a_2+\cdots+a_n}.$$

Due to the inequality from this question $$\sum_{k=1}^{n}\frac{k}{a_1+a_2+\cdots+a_k}<2\sum_{k=1}^{n}\frac{1}{a_k},\qquad n\in\mathbb{N},$$ we can conclude that $$\sum_{n=1}^{\infty}\frac{1}{a_n}<\infty\implies\sum_{n=1}^{\infty}\frac{n}{a_1+a_2+\cdots+a_n}<\infty.$$ $\textbf{What I want to do is to prove or disprove}$: $$\sum_{n=1}^{\infty}\frac{n}{a_1+a_2+\cdots+a_n}<\infty\implies\sum_{n=1}^{\infty}\frac{1}{a_n}<\infty.$$ Some progress: When $a_n<a_{n+1}$, it is easy to see $$\frac{1}{a_n}<\frac{n}{a_1+a_2+\cdots+a_n},$$ which implies that $$\sum_{n=1}^{\infty}\frac{n}{a_1+a_2+\cdots+a_n}<\infty\implies\sum_{n=1}^{\infty}\frac{1}{a_n}<\infty.$$

How about the general case? Any help and hint will welcome!

Especially Lime
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Riemann
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This isn't true. Start from any sequence where $\sum\frac{n}{a_1+\cdots+a_n}$ converges. If you choose any value $k$ and replace $a_k$ by $a_k+a_{k+1}-1$ and $a_{k+1}$ by $1$, you decrease the value of $\frac{k}{a_1+\cdots+a_k}$ without changing any other term, so this still converges. Do this for infinitely many different (non-consecutive) values for $k$ and the sequence $a_n$ has infinitely many terms equal to $1$, so $\sum\frac{1}{a_n}$ doesn't converge.

Especially Lime
  • 46,692