Assume $a_i>0$ and $\sum\limits_{n=1}^\infty\dfrac{1}{a_i}$ is convergent. Show the series $$\sum_{n=1}^\infty\frac{n^2a_n}{(a_1+a_2+\cdots+a_n)^2}$$ is convergent.

Since we can get the convergence of $$\sum_{n=1}^\infty\frac{n}{a_1+a_2+\cdots+a_n},$$ is it feasible to apply Abel's test? Or I should use other methods?

  • $\begingroup$ the convergence of the second series come from the first one. by cauchy Schwartz $\endgroup$ – Guy Fsone Nov 30 '17 at 5:08
  • $\begingroup$ @GuyFsone Sorry, could you explain it in detail? I don't get it. Many thanks! $\endgroup$ – username Nov 30 '17 at 5:42
  • $\begingroup$ I don't see how to get the first inequality $\endgroup$ – Guy Fsone Nov 30 '17 at 11:46

Let $A_n=a_1+\ldots+a_n$ and $A_0=a_1$. We may notice that $$ \sum_{n\geq 1}\frac{n^2 a_n}{A_n^2}=\sum_{n\geq 1}\frac{n^2 (A_n-A_{n-1})}{A_n^2}\leq\sum_{n\geq 1}\frac{n^2 (A_n-A_{n-1})}{A_n A_{n-1}}=\sum_{n\geq 1}n^2\left(\frac{1}{A_{n-1}}-\frac{1}{A_n}\right)$$ and by summation by parts $$\begin{eqnarray*} \sum_{n=1}^{N}n^2\left(\frac{1}{A_{n-1}}-\frac{1}{A_n}\right)&=&N^2\left(\frac{1}{a_1}-\frac{1}{A_N}\right)-\sum_{n=1}^{N-1}(2n+1)\left(\frac{1}{a_1}-\frac{1}{A_n}\right)\\&=&\frac{1}{a_1}-\frac{N^2}{A_N}+\sum_{n=1}^{N-1}\frac{2n+1}{A_n}.\end{eqnarray*}$$ By the Cauchy-Schwarz inequality (in the Titu's lemma form) $$ \sum_{n=1}^{N}\frac{1}{a_n}\geq \frac{N^2}{A_N} $$ hence it is enough to show that $\sum_{n\geq 1}\frac{2n+1}{A_n}$ is convergent. This is granted by Knopp's inequality: see exercise 212 of my notes (page 109): $$ \sum_{n\geq 1}\frac{2n+1}{A_n}\leq 4\sum_{n\geq 1}\frac{1}{a_n}.$$


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