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Suppose that $a_n > 0$ for all $n\geq 1$, and define $S_n = \sum_{i=1}^n{a_i}$. If $S_n$ is convergent, prove that $$ \frac{n^2}{\sum_{i=1}^n{\frac{1}{a_i}}} $$ is also convergent.

Thanks.

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You forgot to say "please". –  Stijn Mar 20 '11 at 9:25

2 Answers 2

up vote 6 down vote accepted

Check out the Wikipedia article on Pythagorean means.

Here are some more details. The Wikipedia article tells us that the quantity in question is bounded by the sum. That appears only to give boundedness. Suppose, however, that $\sum_{i=N}^\infty a_i < \epsilon$, and let $C = \sum_{i=1}^{N-1} a_i^{-1}$. For $n > N$ we have $$\frac{(n-N)^2}{\sum_{i=N}^n a_i^{-1}} < \epsilon.$$ In other words, $$ \sum_{i=N}^n a_i^{-1} > \epsilon^{-1} (n-N)^2. $$ Therefore $$ \sum_{i=1}^n a_i^{-1} > C + \epsilon^{-1} (n-N)^2 = \epsilon^{-1}n^2 + o(n^2). $$ So in fact $$ \limsup_{n \rightarrow \infty} \frac{n^2}{\sum_{i=1}^n a_i^{-1}} < \epsilon. $$ Since this is true for every $\epsilon > 0$, the limit in question is actually $0$.

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Not sure this yields more than the boundedness. –  Did Mar 20 '11 at 13:00
    
@Didier That depends on what you do with it. –  Yuval Filmus Mar 20 '11 at 13:23
    
Indeed, as the (vastly expanded) new version of your answer shows. –  Did Mar 20 '11 at 13:31

This is similar to Yuval Filmus' proof, but perhaps different enough to be worth considering.

Since $\sum_{i=1}^n a_i$ converges, for any $\epsilon>0$ there is an $N$ so that for all $m,n>N$, $$\sum_{i=m+1}^n a_i<\epsilon$$ Holder's inequality says $$ \left(\sum_{i=m+1}^n a_i\right)\left(\sum_{i=m+1}^n \frac{1}{a_i}\right)\ge\left(\sum_{i=m+1}^n 1\right)^2=(n-m)^2 $$ Therefore, for all $m,n>N$, $$ \frac{(n-m)^2}{\displaystyle\sum_{i=1}^n\frac{1}{a_i}}\le\frac{(n-m)^2}{\displaystyle\sum_{i=m+1}^n\frac{1}{a_i}}\le\sum_{i=m+1}^n a_i<\epsilon $$ Let $n\ge 2m>2N$, then $$ \frac{n^2}{\displaystyle\sum_{i=1}^n\frac{1}{a_i}}\le 4\frac{(n-m)^2}{\displaystyle\sum_{i=1}^n\frac{1}{a_i}}<4\epsilon $$ Since $\epsilon$ was arbitrary, we get that $$ \lim_{n\to\infty}\;\frac{n^2}{\displaystyle\sum_{i=1}^n\frac{1}{a_i}}=0 $$

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