How prove this limit $\lim_{n\to\infty}\frac{n^2}{\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{n}}}=0$

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.