sequences and series - Is my proof ok? If $sum u_n$ diverges then $sum frac {u_n} {u_1 + u_2 + dots + u_n}$ also diverges

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.

Answer

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.