Prove the convergence of this sequence I have read the following problem. Let $(a_n)_n$ be a convergent sequence and let $b_n:=\frac{1}{n}\sum_{i=1}^{n}(-1)^{i+1}a_i$ be the accompanying alternating partial sum that is multiplied by $\frac{1}{n}$. Show that the sequence $(b_n)_n$ converges to $0$.
I haven't done problems like these for quite some time, so I couldn't really come up with an answer. The sum to $n$ and $\frac{1}{n}$ made it seem that maybe I could work with some boundedness argument, so I tried something like the following. Since $(a_n)_n$ is convergent, it is bounded by some $c$. Thus we have that $$|b_n|\leq \sum_{i} \frac{1}{n} |a_i| \leq c$$ but this doesn't help me at all it seems, especially not for an epsilon argument.
 A: Fix $\epsilon > 0$ and let $a$ and $n_0$ be such that $n\geq n_0$ implies $|a_n-a| < \epsilon$. Then
$$ \frac{1}{n} \sum_{i=1}^n (-1)^{i+1} a_n = \frac{1}{n} \sum_{i=1}^{n_0}(-1)^{i+1}a_i + \frac{1}{n}\sum_{n_0+1}^{n} (-1)^{i+1} a +  \frac{1}{n}\sum_{n_0+1}^{n} (-1)^{i+1} \epsilon_i, $$
where $\epsilon_i := a_n - a$.
The first sum converges to zero trivially. Now $\sum_{n_0+1}^n(-1)^{i+1}a $ is either $0$ or $\pm a$, so again $\frac{1}{n}\sum_{n_0+1}^n(-1)^{i+1} a$ converges to zero. Finally, $\sum_{n_0+1}^n\epsilon_i$ is at least $-n\epsilon$ and at most $n\epsilon$, so that $\frac{1}{n}\sum_{n_0+1}^{n}(-1)^{i+1} \epsilon_i $ is at least $-\epsilon$ and at most $\epsilon$.
A: There is a general theorem that if $b_k\to b$ then $${1\over n}\sum_{k=1}^nb_k\to b$$ Let $$b_k=(-1)^{k+1}[a_k-a_{k-1}]=(-1)^{k+1}a_k+(-1)^ka_{k-1}$$ where $a_0=0.$ Then $b_k\to 0.$ Hence $${1\over n}\sum_{k=1}^nb_k={2\over n}\sum_{k=1}^n(-1)^{k+1}a_k-{1\over n}(-1)^{n+1}a_n\to 0$$ The last term tends to $0.$ Therefore $${1\over n}\sum_{k=1}^n(-1)^{k+1}a_k\to 0$$
