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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.

Yotam Alon
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  • See this. – David Mitra Feb 02 '15 at 21:16
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    I have tried proving that $\sum S_n$ also diverges and i thought that that would help but there are a few different cases, if $a_n$ converges to a finite number, converges to infinity, or diverges – Yotam Alon Feb 02 '15 at 21:17
  • For context, this is an exercise in Rudin's Introduction to Mathematical Analysis, specifically Exercise 11(b) in the chapter Numerical Sequences and Series. The members of this site are often resistant to explicitly solving a homework problem without seeing effort beforehand. Could you please tell us exactly what you are having trouble with and what you have tried? – Joel Feb 02 '15 at 21:17
  • @david Mitra: looks pretty similar, let me take a closer look – Yotam Alon Feb 02 '15 at 21:18
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    Clarification: Some members of this site are resistant to explicitly solving homework problems. Not all. – Pp.. Feb 02 '15 at 21:19
  • In Rudin, there is a hint: Prove that $$\frac{a_{N+1}}{s_{N+1}} + \cdots + \frac{a_{N+k}}{s_{N+k}} \ge 1 - \frac{s_{N}}{s_{N+k}}$$ – Joel Feb 02 '15 at 21:19
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    Also not quite sure i understood the question the David pointed out... and to be clear i want to be able to solve it on the test, so i don't expect any easy answers. – Yotam Alon Feb 02 '15 at 21:23
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    @YotamAlon This wasn't clear from the post, unfortunately. –  Feb 03 '15 at 00:13
  • @DavidMitra Why didn't you vote to close as a duplicate? – Pedro Feb 03 '15 at 16:02

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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 $\epsilon$.

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.

Joel
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    i didn't quite understand how you concluded from the fact that $S_N+k$ diverges that the tail is greater or equal to 1. can you explain that step please? – Yotam Alon Feb 02 '15 at 22:27