# $a_n > 0$ 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. [duplicate]

$$a_n > 0$$ 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.

• 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.
– kccu
Mar 29, 2019 at 14:54
• See this posting and the links: math.stackexchange.com/questions/2952345/… Mar 29, 2019 at 15:51

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