Divergence of $\sum\limits_{k=1}^n \frac{a_n}{S_n}$ where $S_n = \sum\limits_{k=1}^n a_k$, when $S_n$ diverges [duplicate]

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.

Solution 1:

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.

Solution 2:

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