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$a_n > 0$ and $\sum_\limits{n=1}^{+\infty} \frac{1}{a_n}$ converges. Prove $\sum_\limits{n=1}^{+\infty} \frac{n}{a_1 + \cdots + a_n}$ is convergent.

I find that this may have something to do with Stolz Theorem which says that if $\{\frac{1}{a_n}\}$ is convergent then $$\lim_{n \rightarrow +\infty} \frac{n}{a_1+\cdots +a_n} = \lim_{n \rightarrow +\infty} \frac{1}{a_n}$$ This may implies that $\frac{n}{a_1+\cdots +a_n}$ and $\frac{1}{a_n}$ have the same declining speed so leads to the answer of the question.

However, I don't know how to turn this into correct proof.

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    $\begingroup$ Saying that $\lim_{n \to \infty} \frac{n}{a_1+\cdots+a_n}=\lim_{n\to \infty} \frac{1}{a_n}$ does not imply they have "the same declining speed." E.g., $\lim_{n\to \infty} \frac{1}{n^2}=\lim_{n\to \infty} \frac{1}{n}=0$ but one of these sequences is summable and the other is not. $\endgroup$
    – kccu
    Mar 29 '19 at 14:54
  • $\begingroup$ See this posting and the links: math.stackexchange.com/questions/2952345/… $\endgroup$ Mar 29 '19 at 15:51
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Let $b_n = \frac{1}{a_n}$, then $\frac{n}{a_1 + \cdots + a_n}$ is the harmonic mean of $b_1,..b_n$ so it is less or equal than the corresponding geometric mean.

$\sum_\limits{n=1}^{+\infty} \frac{n}{a_1 + \cdots + a_n} \leq \sum_\limits{n=1}^{+\infty} (b_1....b_n)^{\frac{1}{n}} \leq e\sum_\limits{n=1}^{+\infty} b_n=e\sum_\limits{n=1}^{+\infty} \frac{1}{a_n} < \infty$ by Carleman's Inequality

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