If $a_n\to a$ then $b_n=\frac{1}{n^2}\sum_{i=1}^n ia_i \to a/2$. Suppose the sequence $\{a_n\}$ of real numbers has limit $a$.  Define $b_n=\frac{1}{n^2}\sum_{i=1}^n ia_i$ and prove that $b_n\to a/2$.

My attempts: I first tried writing down the definition of the limit.
$$ \left|\frac{1}{n^2}\sum_{i=1}^n ia_i - a/2\right| $$
I don't think I have any great ideas about what to do here.  I could distribute the $1/n^2$ but that doesn't seem useful.  I could add and subtract terms looking for a way to apply the triangle inequality.  But what terms?  I thought about maybe $\sum_{i=1}^n a_i$ or something like that, but it doesn't seem useful as far as I can tell.
I also thought maybe the AM-GM inequality might be useful.  If I were merely trying to prove convergence and not to some point that might be true, but I don't see how to use it here.
I thought this also kinda looks like something I've seen in Calc 2 where you take a derivative to find a Taylor expansion.  But here there is no polynomial just an arbitrary sequence $\{a_n\}$.
So I'm out of ideas at this point.

Let $\epsilon>0$ be given. By definition, there exists $N\in\mathbb{N}$ such that $n\geq N$ implies $|a_n-a|<\epsilon$. This then implies $a-\epsilon<a<a+\epsilon$ and we have
$$\sum_{i=1}^n i a_i=\sum_{i=1}^N ia_i+\sum_{i=N+1}^n ia_i$$
$$\Rightarrow \sum_{i=N+1}^n i(a-\epsilon)<\sum_{i=1}^n i a_i-\sum_{i=1}^N ia_i<\sum_{i=N+1}^n i(a+\epsilon)$$
$$\Rightarrow (a-\epsilon)\frac{1}{2}(n-N)(n+N+1)<\sum_{i=1}^n i a_i-\sum_{i=1}^N ia_i<(a+\epsilon)\frac{1}{2}(n-N)(n+N+1)$$
This then implies
$$(a-\epsilon)\frac{(n-N)(n+N+1)}{2n^2}<\frac{1}{n^2}\sum_{i=1}^n i a_i-\frac{1}{n^2}\sum_{i=1}^N ia_i<(a+\epsilon)\frac{(n-N)(n+N+1)}{2n^2}$$
Taking limits as $n$ approaches infinity gives
$$\frac{a-\epsilon}{2}=\lim_{n\to\infty}(a-\epsilon)\frac{(n-N)(n+N+1)}{2n^2}$$
$$<\lim_{n\to\infty}b_n+\left[\frac{1}{n^2}\sum_{i=1}^N \right]ia_i\lim_{n\to\infty}\frac{1}{n^2}$$
$$<\lim_{n\to\infty}(a+\epsilon)\frac{(n-N)(n+N+1)}{2n^2}=\frac{a+\epsilon}{2}$$
Since the limit of $\frac{1}{n^2}$ is $0$ this simplifies to
$$\frac{a-\epsilon}{2}<\lim_{n\to\infty}b_n<\frac{a+\epsilon}{2}$$
Since $\epsilon$ was arbitrary we conclude
$$\frac{a}{2}\leq \lim_{n\to\infty}b_n\leq \frac{a}{2}$$
We conclude
$$\lim_{n\to\infty}b_n=\frac{a}{2}$$

EDIT: As requested, if for all $\epsilon>0$ we have
$$a_n-\epsilon<b_n<a_n+\epsilon$$
and $a_n\to a$, then $b_n\to a$. First, note that we can rearrange this to
$$|b_n-a_n|<\epsilon$$
Let $\delta>0$ be given and choose $\epsilon=\frac{\delta}{2}$. Then choose $N$ such that $n\geq N$ implies $|a_n-a|<\frac{\delta}{2}$. This gives us
$$|b_n-a|=|b_n-a+a_n-a_n|\leq |b_n-a_n|+|a-a_n|<\frac{\delta}{2}+\frac{\delta}{2}=\delta$$
as desired.
