<s>[INST] 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.
 [/INST] 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 $(n!)^{1/n} \ge \frac{n+1}{\mathrm{e}}$.
</s>