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The question is :

If $\sum u_n$ is a divergent series of positive real numbers and $s_n = u_1 + u_2 + \dots + u_n$ , prove that the series $\sum \frac {u_n} {s_n}$ is divergent.

I tried my best. But I failed. Please help me by giving me a hint. Thank you in advance.

My solution :

Let $\{t_n\}$ be the sequence of partial sums of the series $\sum \frac {u_n} {u_1 +u_2 + \dots +u_n}$. Let $s_n=u_1 +u_2+ \dots +u_n$. If we can prove that $\{t_n\}$ is not Cauchy then our purpose will be served. For this we have to show that $\exists \epsilon>0$ such that $\forall n \in \mathbb N, \exists p \in \mathbb N$ for which $t_{n+p}-t_n \geq \epsilon$. Now since $\sum u_n$ is divergent, then so is $\{s_n\}$. Hence $\forall n \in \mathbb N, \exists q \in \mathbb N$ such that $s_{n+q}>2s_n$.

Now, $t_{n+q}-t_n= \frac {u_{n+1}} {s_{n+1}} + \frac {u_{n+2}} {s_{n+2}} + \dots + \frac {u_{n+q}} {s_{n+q}} > \frac {u_{n+1} +u_{n+2}+\dots +u_{n+q}} {s_n} > \frac {s_{n+q} -s_n} {s_n} >1$ which proves that $\{t_n\}$ is not a Cauchy sequence, hence the series $\sum \frac {a_n} {s_n}$ is divergent.

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    $\begingroup$ I'm not sure how this can be marked as a duplicate since OP is asking for a critique of his proof. Since he also is specifically looking for a hint, directing him to look at other answers would seem a bad idea. $\endgroup$ – qbert Jul 20 '16 at 7:33
  • $\begingroup$ @qbert does my solution hold good or are there more interesting proofs? If the answer is 'Yes', let me aware of these proofs. It will perhaps useful for concretizing my thoughts and conceptions concerning infinite series. $\endgroup$ – user251057 Jul 20 '16 at 8:23
  • $\begingroup$ @A.Chattopadhyay: My post fully answers your questions in the above comment. $\endgroup$ – Alex M. Jul 20 '16 at 8:28
  • $\begingroup$ @user251057 Why $s_{n+q}>2s_n $ $\endgroup$ – user1942348 May 30 '18 at 7:17
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Your idea is nice, unfortunately its very last step has a crucial mistake: the largest of the numbers $s_{n+1}, \dots, s_{n+q}$ is $s_{n+q}$ (because $\{s_n\}$ increases), so the corect inequality is

$$\frac {u_{n+1}} {s_{n+1}} + \frac {u_{n+2}} {s_{n+2}} + \dots + \frac {u_{n+q}} {s_{n+q}} > \frac {u_{n+1} +u_{n+2}+\dots +u_{n+q}} {\color {red} {s_{n+q}}} > \frac {s_{n+q} -s_n} {\color {red} {s_{n+q}}} = 1 - \frac {s_n} {s_{n+q}}$$

and with this you have reached a dead end, because $1 - \frac {s_n} {s_{n+q}} < 1$, which does not help you.

This question has been asked awfully many times here, just see this instance of it and from there follow the links that take you along a full chain of duplicates, full of various solutions.

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  • $\begingroup$ Yeah you are right @Alex M. $\endgroup$ – user251057 Jul 20 '16 at 8:28
  • $\begingroup$ I just want to show $s_{n+q}>2s_n$ $\endgroup$ – user1942348 May 30 '18 at 11:49

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