# Proving limit implies another limit

I'm working out some study problems that I don't have answers to yet. Could someone help me out with this one please?

Let $a_n$ and $b_n$ be two positive sequences such that $$b_n = \frac{a_1+a_2...+a_n}{n}.$$ Prove that $\lim_{n\to \infty} a_n=0$ implies $\lim_{n\to \infty} b_n=0$. Is the converse true?

Given $\epsilon>0$, we know there is some $N$ such that $a_n < \epsilon/2$ for all $n > N$. Having chosen this $N$, there exists another $N'\ge N$ such that $\frac{a_1+\cdots+a_N}{n} < \epsilon/2$ for all $n \ge N'$. Then for $n>N'$, $$b_n = \frac{a_1+\cdots+a_N}{n} + \frac{a_{N+1}+\cdots+a_n}{n} < \epsilon/2 + \frac{n\epsilon/2}{n} = \epsilon.$$
Converse: consider the sequence $a_n=(-1)^n$. Then $b_n = \begin{cases}0 & \text{$n$even}\\ -1/n & \text{$n$odd}\end{cases}$ and tends to zero, while $a_n$ does not converge.
To prove the desired convergence, you can use the Cauchy criterion; i.e., use the fact that $a_n$ is Cauchy to examine the differences $$b_{m} - b_{n}.$$
The converse is false: take $a_{n} = (-1)^n$.