How prove this limit $\lim_{n\to\infty}\frac{n^2}{\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{n}}}=0$ Assume that a positive term series $\displaystyle\sum_{n=1}^{\infty}a_{n}$ converges,
show that
$$\lim_{n\to\infty}\dfrac{n^2}{\dfrac{1}{a_{1}}+\dfrac{1}{a_{2}}+\cdots+\dfrac{1}{a_{n}}}=0.$$
My try: since $a_{n}>0$. so
$$\dfrac{n^2}{\dfrac{1}{a_{1}}+\dfrac{1}{a_{2}}+\cdots+\dfrac{1}{a_{n}}}>0.$$
But I can't find this right limit is zero!.Thank you.
 A: Let  $\varepsilon >0$.
There is an index $r$ such that $\sum_{k=r}^{\infty}a_k \leq \frac{4}{9}\varepsilon$.
Let $n\geq r$. Consider the $n-r+1$ numbers $a_r,a_{r+1}, \ldots,a_n$. The
generalized means inequality 
implies that
$$
M_{-1}(a_r,a_{r+1}, \ldots,a_n) \leq M_{0}(a_r,a_{r+1}, \ldots,a_n) \tag{1}
$$ 
i.e.
$$
\frac{n-r+1}{\sum_{k=r}^{n} \frac{1}{a_k}} \leq \frac{\sum_{k=r}^n a_k}{n-r+1} \tag{2}
$$
This implies that
$$
\frac{n^2}{\sum_{k=1}^{n} \frac{1}{a_k}} \leq \frac{n^2}{\sum_{k=r}^{n} \frac{1}{a_k}} \leq
\big(\frac{n}{n-r+1}\big)^2 \big(\sum_{k=r}^n a_k \big) \leq
\big(\frac{n}{n-r+1}\big)^2 \frac{4}{9}\varepsilon
$$
Now, if we take $n\geq 3(r-1)$ we will have $\frac{n}{n-r+1} \leq \frac{3}{2}$
so $\big(\frac{n}{n-r+1}\big)^2 \leq \frac{9}{4}$ and hence
$$
\frac{n^2}{\sum_{k=1}^{n} \frac{1}{a_k}} \leq \varepsilon
$$
which concludes the proof.
A: Here is a far from rigorous argument.  Because 
$$\sum_{n=1}^{\infty} a_n < \infty$$
this implies that $a_n < (1/C) n^{-1-\epsilon}$ $\forall \, \epsilon > 0$ and some $C>0$.  Therefore $1/a_n >C  n^{1+\epsilon}$.  By the integral test,
$$\sum_{n=1}^{\infty} \frac{1}{a_n} \gt C' n^{2+\epsilon}$$
for $C' >0$.  Therefore, the expression in the limit behaves as $(1/C') n^{-\epsilon}$ as $n \to \infty$, and the limit is zero.
A: It suffices to find some bounds $C(n, \{a_k\}_{k=1}^n)$ with $\lim_{n\to \infty} C(n, \{a_k\}_{k=1}^n) = 0$ such that 
$$\frac{n^2}{\frac{1}{a_1} + \frac{1}{a_2} + \cdots + \frac{1}{a_n}}
\le C(n, \{a_k\}_{k=1}^n).$$
I knew two bounds as follows. The proof of $\lim_{n\to \infty} C(n, \{a_k\}_{k=1}^n) = 0$ is easy and thus omitted.
1) We have
$$\frac{n^2}{\frac{1}{a_1} + \frac{1}{a_2} + \cdots + \frac{1}{a_n}}
\le 4\sum_{k = \lfloor \frac{n}{2}\rfloor}^n a_k.$$
Remark: This solution is the same as @Ewan Delanoy's one.
2) We have
$$\frac{n^2}{\frac{1}{a_1} + \frac{1}{a_2} + \cdots + \frac{1}{a_n}}
\le \frac{\mathrm{e}}{n+1}\sum_{k=1}^n ka_k.
$$
$\phantom{2}$
Some proofs:
For the first bound: By Cauchy-Bunyakovsky-Schwarz inequality, we have
\begin{align}
\frac{n^2}{\frac{1}{a_1} + \frac{1}{a_2} + \cdots + \frac{1}{a_n}}
&\le \frac{n^2}{\frac{1}{a_{\lfloor\frac{n}{2}\rfloor}} + \frac{1}{a_{\lfloor\frac{n}{2}\rfloor+1}} + \cdots + \frac{1}{a_n}}\\
&\le \frac{n^2}{(n - \lfloor\frac{n}{2}\rfloor + 1)^2}
(a_{\lfloor\frac{n}{2}\rfloor} + a_{\lfloor\frac{n}{2}\rfloor + 1} + \cdots + a_n)\\
&\le 4\sum_{k = \lfloor \frac{n}{2}\rfloor}^n a_k.
\end{align}
For the second bound: By using GM-HM and AM-GM, we have
\begin{align}
\frac{n^2}{\frac{1}{a_1} + \frac{1}{a_2} + \cdots + \frac{1}{a_n}}
&\le n(a_1a_2\cdots a_n)^{1/n}\\
&= \frac{n}{(n!)^{1/n}} (a_1\cdot 2a_2 \cdot 3a_3 \cdots na_n)^{1/n}\\
&\le \frac{1}{(n!)^{1/n}} \sum_{k=1}^n ka_k\\
&\le \frac{\mathrm{e}}{n+1}\sum_{k=1}^n ka_k
\end{align}
where we have used the Stirling approximation to get $(n!)^{1/n} \ge \frac{n+1}{\mathrm{e}}$.
