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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.

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marked as duplicate by Pedro Tamaroff Feb 3 '15 at 16:01

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  • $\begingroup$ See this. $\endgroup$ – David Mitra Feb 2 '15 at 21:16
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    $\begingroup$ 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 $\endgroup$ – Yotam Alon Feb 2 '15 at 21:17
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    $\begingroup$ Clarification: Some members of this site are resistant to explicitly solving homework problems. Not all. $\endgroup$ – Pp.. Feb 2 '15 at 21:19
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    $\begingroup$ 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. $\endgroup$ – Yotam Alon Feb 2 '15 at 21:23
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    $\begingroup$ @YotamAlon This wasn't clear from the post, unfortunately. $\endgroup$ – user147263 Feb 3 '15 at 0:13
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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.

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    $\begingroup$ 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? $\endgroup$ – Yotam Alon Feb 2 '15 at 22:27

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