Prove that $ \frac{n^2}{\sum_1^n{\frac{1}{a_i}}} $ is convergent. 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.
 A: 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 $0$.
A: 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
$$
\lim_{n\to\infty}\;\frac{n^2}{\displaystyle\sum_{i=1}^n\frac{1}{a_i}}=0
$$
