Proving convergence of series with comparison test.

Question

Let $S_n = a_1 + \ldots + a_n$ where $a_n > 0$ for all $n$. I'm trying to prove that $\sum_{n=1}^\infty a_n$ converges if and only if $\sum_{n=1}^\infty \frac{a_n}{S_n}$ converges using a comparison test.

I succeeded for the if part: If $\sum_{n=1}^\infty a_n = \lim_{n \to \infty}S_n = S$ then $S> 0$ and $S_n > S/2$ for all $n > n_0$ and $a_n/S_n < (2/S)a_n$. By comparison test it holds that $\sum_{n=1}^\infty \frac{a_n}{S_n}$ converges.

For the only if part I am having problems. Assuming $\sum_{n=1}^\infty \frac{a_n}{S_n}$ converges I don't know that $a_n < a_n/S_n$ because there is no restriction that $S_n > 1$. The limit comparison test fails because I don't know that $a_n/(a_n/S_n) = S_n$ converges yet either. Any help is appreciated.

Answer 1

For a proof of the reverse implication using comparison, assume that $\sum a_n$ diverges. We have $S_n \to \infty$ since the terms are positive and note that $\displaystyle\frac{a_n}{S_{n-1}} = \frac{a_n}{a_n + S_{n-1}}$.

We either have (1) $a_n \leqslant S_{n-1}$ for all sufficiently large $n > N$ or (2) there exists a subsequence $a_{n_k}$ such that $a_{n_k} > S_{n_k-1}$.

In case (2) we have for infinitely many $n_k$,$$\frac{a_{n_k}}{S_{n_k}} \geqslant \frac{a_{n_k}}{2 a_{n_k}} = \frac{1}{2}$$

and the series $\sum \frac{a_n}{S_n}$ clearly diverges.

In case (1) we have for all $n > N$,

$$\frac{a_{n}}{S_{n}} \geqslant \frac{a_n}{2s_{n-1}}= \frac{S_n - S_{n-1}}{2S_{n-1}}> \int_{S_{n-1}}^{S_n} \frac{dx}{2x}$$

and since $S_n \to \infty$ the series diverges by the (integral) comparison test.