Let $a_n$ be a positive sequence such that $S_n = \sum\limits_{k=1}^n a_k$ diverges. I'm trying to prove $\sum\limits_{k=1}^n \frac{a_n}{S_n}$ diverges.

I tried summation by parts, limit comparison and Stolz theorem in many different combinations and am still stuck with nothing.


marked as duplicate by user147263, Tim Raczkowski, Willie Wong, Claude Leibovici, daw Dec 1 '15 at 8:51

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Note that, for every $M\geqslant N$, $$\sum_{n=N+1}^M\frac{a_n}{S_n}\geqslant\sum_{n=N+1}^M\frac{a_n}{S_M}=1-\frac{S_N}{S_M}$$ hence, for every $N$, considering that $S_M\to+\infty$ when $M\to\infty$, $$ \sum_{n=N+1}^\infty\frac{a_n}{S_n}\geqslant1. $$ This shows that the series diverges since the rest of a summable series converges to zero.

  • 1
    $\begingroup$ Elegant solution. $\endgroup$ – Olivier Bégassat May 25 '14 at 16:27
  • $\begingroup$ Instead of "rest" I would say "remainder." $\endgroup$ – Pedro Tamaroff Feb 3 '15 at 16:04

HINT : If $\lim_{n\rightarrow +\infty} \frac{a_n}{S_n}=0$ $$ \frac{a_n}{S_n}\sim -\ln(1-\frac{a_n}{S_n})=\ln(\frac{S_n}{S_{n-1}}) $$

Details :

$$\sum_{p=1}^{n}\ln(\frac{S_p}{S_{p-1}})=\sum_{p=1}^n \ln(S_p)-\sum_{p=0}^{n-1}\ln(S_p)=\ln(S_n)-\ln{S_0}$$

As the series $\sum a_n$ diverges then the series $\sum\ln(\frac{S_n}{S_{n-1}})$ diverges and therefore $\sum \frac{a_n}{S_n}$

  • $\begingroup$ Nice approach (although the $\sim$ step needs to be made more rigorous). $\endgroup$ – Did May 25 '14 at 16:35
  • $\begingroup$ @Did You are right. But initially it was only a 'trick' to help user136640 which I hope he can do it by itself. $\endgroup$ – Krokop May 25 '14 at 16:48
  • $\begingroup$ I don't think you can make this approach work, because there is no reason to expect $\frac{a_n}{S_n}$ to be close to $0$. It could very well tend to $1$ for instance, in which case $-\ln\left(1-\left(\frac{a_n}{S_n}\right)\right)$ tends to infinity. $\endgroup$ – Olivier Bégassat May 25 '14 at 17:19
  • $\begingroup$ @OlivierBégassat We are dealing with two cases: first one -we do not have $\lim_{n\rightarrow +\infty} \frac{a_n}{S_n}=0$ and the second case where the limit is $0$. $\endgroup$ – Krokop May 25 '14 at 17:21
  • $\begingroup$ Missed that, my bad!${}{}{}$ $\endgroup$ – Olivier Bégassat May 25 '14 at 17:34

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