How to prove that $\sum_{n}^{\infty}u_{n}^{2}\left(u_{1}+\cdots+u_{n}\right)^{-1}<\infty$ under these conditions?

Question

Suppose that $u_n$ is a decreasing sequence of positive numbers that converges to zero. Suppose moreover that $S_n = \sum_{k=1}^n u_k$ diverges. I would like to prove that the sum $\sum u_k^2 / S_k$ is finite, which I believe is true. Indeed, sum of the type $\sum u_k / S_k^{1+\epsilon}$ are finite, but I have not been able to exploit that. Comparison with an integral of the type $\int (f')^2(x)/f(x) \, dx$ has not revealed particularly helpful to gain any intuition. Have also tried to prove that this is a Cauchy sequence, but I always end up missing something.

Thank you.

Answer 1

Hint: use that $S_n \geq nu_n$ for every $n \in \mathbb{N}$. We then have $\frac{u_n^2}{S_n} \leq \frac{u_n}{n}$

If we call $U_n \overset{\Delta}{=} \sum_{i=1}^n \frac{u_k^2}{S_k}$, what can you say about the general term of the sum?

Answer 2

Consider a sequence that really slowly tends to $0$. For example

$$u_n = \frac{1}{\sqrt{\log (n+1)}}.$$

Then you have

$$\frac{n}{\sqrt{\log (n+1)}} \leqslant S_n \leqslant \frac{n}{\sqrt{\log 2}},$$

and

$$\frac{\sqrt{\log 2}}{n\log (n+1)} \leqslant \frac{u_n^2}{S_n} \leqslant \frac{1}{n\sqrt{\log (n+1)}},$$

so in that case

$$\sum_{n=1}^\infty \frac{u_n^2}{S_n} = +\infty.$$

Generally, you have $S_n \leqslant n u_1$, and hence

$$\frac{u_n^2}{S_n} \geqslant \frac{u_n^2}{nu_1},$$

so it is a necessary condition that $\sum \frac{u_n^2}{n} < \infty$. That condition is however not sufficient, as $u_n = (\log (n+1))^{-t}$ for $\frac{1}{2} < t \leqslant 1$ shows, in that case $S_n \sim \frac{n}{(\log (n+1))^t}$ and $\frac{u_n^2}{S_n} \sim \frac{1}{n(\log(n+1))^t}$. On the other hand, we have $S_n \geqslant n u_n$, so the condition $\sum \frac{u_n}{n} < \infty$ is sufficient.