Suppose $\sum_{n=1}^\infty \frac{1}{a_n} = A$ is summable, with $a_n > 0,$ $n = 1,2,3,\cdots.$ How can we prove that $\sum_{n=1}^\infty \frac{n}{a_1 + \dots + a_n}$ is also summable?

This question came from a problem-solving seminar, but I'm quite stuck without a push in the right direction. I tried a few things, including Cauchy-Schwarz (which says $\sum_{n=1}^\infty \frac{n}{a_1 + \dots + a_n} < \sum_{n=1}^\infty \frac{A}{n}$) and also the idea of assuming the latter series diverges and attempting to deduce the divergence of the former series from that, using facts such as $\sum a_n = \infty \implies \sum \frac{a_n}{a_1 + \cdots + a_n} = \infty$. Nothing has worked so far.


The easiest way is to use Carleman's Inequality, we have $$ \root{n}\of{a_1\cdots a_n}\leq \frac{a_1+\cdots+a_n}{n} $$ So $$ \sum_{n=1}^\infty \frac{n}{a_1+\cdots+a_n}\leq \sum_{n=1}^\infty \root{n}\of{\frac{1}{a_1}\cdots \frac{1}{a_n}} \leq e\sum_{n=1}^\infty \frac{1}{a_n}. $$ So the convergence of $\sum\dfrac{1}{a_n}$ implies that of $\sum\dfrac{n}{a_1+\cdots +a_n}.$


Here I propose an other proof that yields the best constant. Using Cauchy-Schwarz inequality we have: $$\sum_{j=1}^k j=\sum_{j=1}^k\sqrt{j^2a_j}\cdot{1\over\sqrt{a_j}}\leq \sqrt{\sum_{j=1}^k j^2a_j}\times\sqrt{\sum_{j=1}^k{1\over a_j}}$$ then $${1\over\sum_{j=1}^k{1/ a_j}}\leq{4\over{k^2(k+1)^2}} \left(\sum_{j=1}^kj^2a_j\right)$$ and this can be written in the following way $${k\over\sum_{j=1}^k{1/ a_j}}\leq{4k\over{2k+1}} \left({1\over k^2}-{1\over (k+1)^2}\right)\left(\sum_{j=1}^kj^2a_j\right)$$ which implies $${k\over\sum_{j=1}^k{1/ a_j}}< 2 \left({1\over k^2}-{1\over (k+1)^2}\right)\left(\sum_{j=1}^kj^2a_j\right)$$ and can be writen in the following way $${k\over\sum_{j=1}^k{1/ a_j}}< 2a_k+A_k-A_{k+1}$$ with $A_k={2\over k^2}\sum_{j=1}^{k-1}j^2a_j$ and $ A_1=0$. Summing these inequalities for $1\leq k\leq n$ we find $$\sum_{k=1}^n{k\over\sum_{j=1}^k{1/ a_j}}< 2\left(\sum_{k=1}^n a_k\right)-A_{n+1}, $$ and this proves the desired inequality since $A_{n+1}\geq0$.

On the other hand, if $c$ is the least constant such that if $n\geq 1$ and $a_1,\ldots,a_n>0$ we have $$\sum_{k=1}^n{k\over\sum_{j=1}^k1/a_j}\leq c \sum_{k=1}^na_k$$ then, by choosing $a_n={1\over n}$ we find $$\sum_{k=1}^n{2\over k+1}\leq c \sum_{k=1}^n{1\over k}$$ that is $2\leq c+1/\sum_{k=1}^n{1\over k}$, and this proves that $2\leq c$ since $\lim_{n\to\infty}\sum_{k=1}^n{1\over k}=+\infty$.

  • $\begingroup$ (+1) This proof is a (beautiful) hybrid of the one I gave above with telescoping and the proof given here. Many thanks for sharing! $\endgroup$ – Chris Oct 19 '14 at 21:15

Allow me to present a short, "magical" solution by thoughtful application of Cauchy-Schwarz.

Define $A_n := a_1 + \cdots + a_n.$ We will first prove the convergence of $\sum_{n=1}^\infty \frac{n^2}{A_n^2}a_n$. Now,

\begin{eqnarray} \sum_{n=1}^N \frac{n^2}{A_n^2}a_n &<& \sum_{n=2}^N \frac{n^2}{A_nA_{n-1}}a_n + \frac{1}{{a_1}} \\ &=& \sum_{n=2}^N \frac{a_n}{A_nA_{n-1}}n^2 + \frac{1}{{a_1}} \\ &=& \sum_{n=2}^N \left( \frac{1}{A_{n-1}} - \frac{1}{A_n}\right)n^2 + \frac{1}{{a_1}} \\ &<& \sum_{n=1}^N \frac{(n+1)^2 - n^2}{A_n} + \frac{2}{a_1} \\ &=& \sum_{n=1}^N \frac{2n + 1}{A_n} + \frac{2}{a_1} = 2\sum_{n=1}^N \frac{n}{A_n} + C \end{eqnarray}

where in the last step we added the sum $\sum \frac{1}{A_n}$ to the constant term. Now,

$$\left(\sum_{n=1}^N \frac{n}{A_n}\right) \le {\left(\sum_{n=1}^N \frac{n^2}{A_n^2}a_n\right)}^{1/2}{\left(\sum_{n=1}^N \frac{1}{a_n} \right)}^{1/2} $$

by Cauchy-Schwarz, so, writing $S_N = \sum_{n=1}^\infty \frac{n^2}{A_n^2}a_n$, we will have

\begin{equation} S_N \le B \sqrt{S_N} + C \end{equation}

for all $N$. This implies convergence of $S_N$, since $\sqrt{S_N} \le B + \frac{C}{\sqrt{S_N}}$ could not hold for large $N$ if $S_N$ diverged.

From the Cauchy-Schwarz inequality derived above, we can thus immediately infer the convergence of our original series $\sum_{n=1}^\infty \frac{n}{A_n}$.


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