Let $\{a_n\}$ be a positive sequence and $\sum_{n=1}^\infty a_n^2=A$ . Calculate $\lim_{N\rightarrow\infty}\frac{1}{\sqrt{N}}\sum_{n=1}^N a_n.$ Let $\{a_n\}$ be a positive sequence and $\sum_{n=1}^\infty a_n^2=A$ . Calculate $$\lim_{N\rightarrow\infty}\frac{1}{\sqrt{N}}\sum_{n=1}^N a_n.$$
I have found that $$\Big(\frac{1}{\sqrt{N}}\sum_{n=1}^N a_n\Big)^2=\Big(\sum_{n=1}^N \frac{a_n}{\sqrt{N}}\Big)^2\leqslant\Big(\sum_{n=1}^N\frac{1}{N}  \Big)\Big( \sum_{n=1}^N a_n^2 \Big)=\sum_{n=1}^N a_n^2 \underset{N\rightarrow\infty}{\longrightarrow} A $$
by Cauchy Inequalty. But if I assume $a_n=\frac{1}{n}$ and $A=\frac{\pi^2}{6}$, then $\lim_{N\rightarrow\infty}\frac{1}{\sqrt{N}}\sum_{n=1}^N a_n=0$ is not $A$. I cannot find a way to prove the limit is $0$.
Thank you for helping!
 A: Motivation: If it is correct that $\sum_{n=1}^\infty a_n^2 = A < \infty $ implies $\lim_{N\to\infty}\frac{1}{\sqrt{N}}\sum_{n=1}^N a_n=0$, then the result does not depend on the actual value $A$, only on the fact that the series converges, and that is equivalent to the fact that the series remainder $\sum_{n=m+1}^\infty a_n^2$ converge to zero for $m \to \infty$.
This idea leads to the following proof:
For $1 \le m < N$ we split the sum in two parts
$$
\frac{1}{\sqrt{N}}\sum_{n=1}^N a_n = \frac{1}{\sqrt{N}}\sum_{n=1}^m a_n + \frac{1}{\sqrt{N}}\sum_{n=m+1}^N a_n
$$
and apply the Cauchy-Schwarz inequality only to the second part:
$$
\frac{1}{\sqrt{N}}\sum_{n=1}^N a_n \le \frac{1}{\sqrt{N}}\sum_{n=1}^m a_n 
+ \sqrt{\frac{N-m}{N} \sum_{n=m+1}^N a_n^2 } \, .
$$
Now fix $m$ and take the $\limsup$ for $N \to \infty$:
$$
 \limsup_{N \to \infty} \frac{1}{\sqrt{N}}\sum_{n=1}^N a_n \le  \sqrt{ \sum_{n=m+1}^\infty a_n^2 } \, .
$$
This holds for all $m$. The right-hand side converges to zero for $m \to \infty$ because the series $\sum_{n=1}^\infty a_n^2$ converges. It follows that
$$
  \lim_{N \to \infty} \frac{1}{\sqrt{N}}\sum_{n=1}^N a_n = 0 \, .
$$
