The last problem I posted had a wrong statement. I recovered the correct one.

Let $(a_n)$ be a sequence of positive real numbers.

Prove that $\sum_{n\geq 0}a_n$ converges iff $\displaystyle \sum_{n\geq 0} \frac{a_n}{\sum_{k=0}^n a_k}$ converges

The direct statement is easy to prove using comparison test.

I'm (again!) stuck with the converse.

I tried summation by part, without success.

Note that the convergence of $\displaystyle \sum_{n\geq 0} \frac{a_n}{\sum_{k=0}^n a_k}$ implies $\displaystyle \frac{a_n}{\sum_{k=0}^n a_k} \to 0$

I don't know what to do next ...


Call $S_n=\sum_{k=0}^n a_k$ and suppose $\sum a_n$ does not converge, hence $\lim S_n = +\infty$. Notice also $(S_n)$ is increasing. Then $\sum_{n=p+1}^q \frac{a_n}{S_n} \geq \sum_{n=p+1}^q \frac{a_n}{S_q} =\frac{S_q-S_p}{S_q}=1-\frac{S_p}{S_q}$.

If $\sum \frac{a_n}{S_n}$ converges, letting $q$ tend to infinity gives: $\sum_{n=p+1}^{+\infty} \frac{a_n}{S_n}\geq 1$, which is absurd, as the rest of a convergent series should have limit 0.

  • $\begingroup$ This is slick. Thank you. $\endgroup$ – Gabriel Romon Jun 4 '14 at 17:27

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