Show that the series $\sum_{n=1}^\infty \frac{a_n}{S_n}$ diverges

Question

Given that the series $\sum_{n=1}^\infty a_n$ of positive terms diverges and that $S_n = a_1 + a_2 + ... + a_n$ prove that $\sum_{n=1}^{\infty} \frac{a_n}{S_n}$ also diverges.

Answer 1

Once you establish the hint in the comments, note that given any $N>0$ the fraction $s_N/s_{N+k} \to 0$ as $k\to \infty$. This is because $s_N$ is held constant and $s_{N+k}$ diverges (since $s_N$ diverges.)

Thus the tail $\sum_{n=N}^\infty a_n/s_n$ is greater than or equal to 1 for all $N$. Hence the series cannot converge, since for any $\epsilon>0$ there must be an $N$ for which the tail is less than $\epsilon$.

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Let's call the partial sums of this series $S_N$.

If a series converges, that means its partial sums converge. This means that the partial sums form a Cauchy sequence. Thus for every $\epsilon >0$ (and let's suppose also that $\epsilon < 1$) there is an $N \in \mathbb{N}$ such that $|S_{n+k} - S_{n}|<\epsilon$ for all $n \ge N$ and $k \in \mathbb{N}$.

The term in the absolute values can be rewritten as $S_{n+k}-S_n = a_{n+1}/s_{n+1} + \cdots + a_{n+k}/s_{n+k}$. This also holds when we let $n=N$.

Now if this series converged we have for sufficiently large $N$: $$\frac{a_{N+1}}{s_{N+1}} + \cdots + \frac{a_{N+k}}{s_{N+k}} < \epsilon < 1$$ for all $k \in \mathbb{N}$.

However, since we have established $$1-\frac{s_{N}}{s_{N+k}} < \frac{a_{N+1}}{s_{N+1}} + \cdots + \frac{a_{N+k}}{s_{N+k}}$$ and $\frac{s_{N}}{s_{N+k}}\to 0$ as $k \to \infty$ we have a contradiction.