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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.

Did
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user136640
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2 Answers2

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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.

Did
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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 $\sum \frac{a_n}{S_n}$

Krokop
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  • Nice approach (although the $\sim$ step needs to be made more rigorous). – Did May 25 '14 at 16:35
  • @Did You are right. But initially it was only a 'trick' to help user136640 which I hope he can do it by itself. – Krokop May 25 '14 at 16:48
  • 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. – Olivier Bégassat May 25 '14 at 17:19
  • @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 $0$. – Krokop May 25 '14 at 17:21
  • Missed that, my bad!${}{}{}$ – Olivier Bégassat May 25 '14 at 17:34