If $\sum a_n^2 $ converges, show that $\lim_{n\to\infty}b_n=0$ Let $(a_n)_n \subset R$ and let $b_n=\dfrac{a_1+a_2+...+a_n}{n}$. Prove that if $\sum_{n=1}^{\infty} a_n^2 $ converges, then $\lim\limits_{n\to\infty} b_n=0$.
From the hypothesis, $\sum_{n=1}^{\infty} a_n^2 $ converges, we get $\lim a_n=0$. Then I have no idea for this problem. Can anyone help me please? Thanks in advanced.
 A: By the Cauchy-Schwarz inequality,
$$
\biggl|\sum_{k=1}^na_k\biggr|\le\biggl(\sum_{k=1}^na_k^2\biggr)^{1/2}\biggl(\sum_{k=1}^n1\biggr)^{1/2}=\biggl(\sum_{k=1}^na_k^2\biggr)^{1/2}n^{1/2}.
$$
Hence,
$$
|b_n|=\biggl|\frac{1}{n}\sum_{k=1}^na_k\biggr|\le\frac1{\sqrt n}\biggl(\sum_{k=1}^na_k^2\biggr)^{1/2}\to0
$$
as $n\to\infty$.
A: 
If $\lim_{n\to\infty}a_n=a$, then
  $$
\lim_{n\to\infty}c_n=a,
$$
  where $(c_n)$ is the sequence of the Cesàro means of the sequence $(a_n)$ defined by
  $$
c_n=\frac{1}{n}\sum_{i=1}^na_i
$$
  for each $n\ge1$.

For every $\varepsilon>0$, there exists $N(\varepsilon/2)\ge1$ such that that
$$
|a_n-a|<\varepsilon/2
$$
when $n>N$.
For $n>N$, we have that
$$
\biggl|\frac{1}{n}\sum_{i=1}^na_i-a\biggr|
 \le\biggl|\frac{1}{n}\sum_{i=1}^{N}(a_i-a)\biggr|+\biggl|\frac{1}{n}\sum_{i=N+1}^n(a_i-a)\biggr|.
$$
Let us choose $n>N$ such that
$$
\frac{N\max_{1\le i\le N}|a_i-a|}n<\varepsilon/2.
$$
Then
$$
\biggl|\frac{1}{n}\sum_{i=1}^{N}(a_i-a)\biggr|\le\frac{N\max_{1\le i\le N}|a_i-a|}n<\varepsilon/2.
$$
Also,
$$
\biggl|\frac1n\sum_{i=N+1}^n(a_i-a)\biggr|\le\frac{n-N}{n}\max_{N+1\le i\le n}|a_i-a|\le\varepsilon/2.
$$
This completes the proof.
Hence, you only need the fact that $a_n\to0$ as $n\to\infty$ to show that $n^{-1}\sum_{k=1}^na_k\to0$ as $n\to\infty$. The square summability of $(a_n)$ is superfluous.
A: Hint 
Let $\varepsilon>0$. There is $N$ s.t. $|a_n|<\varepsilon$ when $n\geq N$. Now, if $n\geq N$,
$$|b_n|=\left|\frac{a_1+...+a_{N-1}}{n}+\frac{1}{n}\sum_{k=N}^n a_k\right|\leq \frac{|a_1|+...+|a_{N-1}|}{n}+\frac{1}{n}\sum_{k=N}^n|a_n|.$$
I let you conclude.
