Convergence of one series If $\lambda_1, \cdots, \lambda_n, \cdots$ are positive real numbers such that $\displaystyle\sum_{k=1}^{\infty}(\lambda_1+\cdots+\lambda_k)=+\infty$, we can affirm that $\displaystyle\sum_{k=1}^{\infty}\frac{\lambda_k}{(\lambda_1+\cdots+\lambda_k)^2}$ converges?
I try use some comparison test, but I have not success. Can someone help me?
Thanks.
 A: Yes, it is true. Let $S_k\stackrel{\rm def}{=} \sum_{j=1}^k \lambda_j$.
By assumption, $\lim_{k\to\infty} S_k = \infty$.
Now,
$$
\sum_{k=1}^\infty \frac{\lambda_k}{S_k^2} 
= \sum_{k=1}^\infty \frac{S_k-S_{k-1}}{S_k^2}
$$
which "should" remind you of something (the discrete analogue) of $\int\frac{f'}{f^2}$. This is no coincidence, and one can make this formal via Abel's theorem:
$$
\sum_{k=1}^\infty \frac{\lambda_k}{S_k^2} 
= \sum_{k=1}^\infty \frac{1}{S_k^2}\cdot (S_k-S_{k-1}).
$$
We know that 


*

*$\sum_k (S_k-S_{k-1})$ is a convergent series (it is a telescopic series, and $\lim_{k\to\infty} S_k = \infty$).

*The sequence $(\frac{1}{S_k^2})_k$ is monotone decreasing (as $(S_k)_k$ is monotone increasing: indeed, the $\lambda_k$'s are positive)

*The sequence $(\frac{1}{S_k^2})_k$ is bounded (as it converges; or, equivalently, is monotone decreasing and positive).
Then the series
$
\sum_{k} \frac{1}{S_k^2}\cdot (S_k-S_{k-1})
$
is convergent by Abel's test.
A: If the $\lambda$'s are all positive, no other hypothesis is needed. Write $S_k:=\lambda_1+\cdots+\lambda_k$. For $k\ge2$ we have
$$0<\frac{\lambda_k}{S_k^2}\le\frac{S_k-S_{k-1}}{S_{k-1}S_k}=\frac1{S_{k-1}}-\frac1{S_k}.$$
By telescoping, the sum of the first $n$ terms satisfies
$$
\sum_{k=1}^n\frac{\lambda_k}{S_k^2}\le\frac{\lambda_1}{S_1^2}+ \left(\frac1{S_1}-\frac1{S_n}\right)\le\frac2{\lambda_1}.$$ So the partial sums are increasing and bounded above, hence the series in question converges. 
