Limit of a sequence with some property Given
$$ a_n >0 \text{ and } \lim_{n \to + \infty} a_n \left( \sum_{k=1}^n a_k \right) =2$$
I need to show that
$$\lim_{n \to + \infty} \sqrt{n}\ a_n=1$$ 
I tried first to compute $$\lim_{n \to + \infty} a_n,$$
but I don't know how, or how to handle these kind of questions so I appreciate any  help.
 A: Let $S_n=a_1+\cdots+a_n$. Then $T_n=S_n^2-S_{n-1}^2=S_n^2-(S_n-a_n)^2=2a_n S_n-a_n^2$. Hence as $a_n\to 0$ by @Clement C comment, $T_n\to 4$. By Cesaro's theorem, we get that $\displaystyle \frac{T_1+\cdots+T_n}{n} \to 4$, hence $\frac{S_n^2}{n}\to 4$, and it is easy to conclude.
A: *

*First, we do have $a_n\xrightarrow[n\to\infty]{} 0$: indeed, by contradiction, suppose not. Then the series $\sum_n a_n$ is not convergent (why?). But as this is a series of non-negative terms, this means $A_N=\sum_{n=1}^N a_n \xrightarrow[N\to\infty]{} \infty$. But since $a_n A_n \xrightarrow[n\to\infty]{} 2$, we then get that $a_n\sim_{n\to\infty} \frac{2}{A_n} \xrightarrow[n\to\infty]{} 0$, contradicting the assumption.

*Now, write $A_N = \sum_{n=1}^N a_n$ with the convention $A_0=0$. By summing the above, the fact that $a_n A_n \xrightarrow[n\to\infty]{} 2$, and properties of divergent series, we have that 
$$
\sum_{n=1}^N a_n A_n \operatorname*{\sim}_{n\to\infty} 2n.
$$
and similarly
$$
\sum_{n=1}^N a_{n+1} A_n \operatorname*{\sim}_{n\to\infty} 2n.
$$
(the latter as $a_{n+1} A_n = a_{n+1} A_{n+1} - a^2_{n+1} \xrightarrow[n\to\infty]{} 2$ as well.) We can now rewrite this as
$$\begin{align*}
\sum_{n=1}^N a_n A_n &= \sum_{n=1}^N (A_n-A_{n-1}) A_n
= \sum_{n=1}^N A_n A_n - \sum_{n=1}^N A_{n-1} A_n
= \sum_{n=1}^N A_n A_n - \sum_{n=0}^{N-1} A_{n} A_{n+1} \\
&= A_N^2 + \sum_{n=1}^{N-1} A_n A_n - \sum_{n=1}^{N-1} A_{n} A_{n+1}
=  A_N^2 + \sum_{n=1}^{N-1} A_n(A_n-A_{n+1})
=  A_N^2 - \sum_{n=1}^{N-1} A_n a_{n+1}
\end{align*}$$
and rearranging,
$$
A_N^2 = \sum_{n=1}^{N-1} A_n a_{n+1} + \sum_{n=1}^N a_n A_n \operatorname*{\sim}_{n\to\infty} 4n.
$$
This implies $A_N\operatorname*{\sim}_{n\to\infty} 2\sqrt{n}$, and (finally!) 
$$
a_n\operatorname*{\sim}_{n\to\infty} \frac{2}{2\sqrt{n}} = \frac{1}{\sqrt{n}}$$
i.e.
$$
\sqrt{n} a_n \xrightarrow[n\to\infty]{}1.
$$
