Assume $a_n \to a$. Then Show that $$\lim_{n \to \infty}\sum_{i=1}^n \frac{((n+1)-i)a_i}{n^2} = \frac{a}{2} $$
So as $a_n \to a$ we know $\forall \epsilon, \exists N, \forall n \ge N ,|a_n - a | < \epsilon $.
Further, $\exists M > 0 $ such that $|a_n| < M, \forall n$.
I want to do something similar to the proof for Cesaro means but it doesn't quite work out. We have:
$\displaystyle\lim_{n \to \infty }\bigg | \frac{na_1 + (n-1)a_2 + ... +2a_{n-1} + a_n}{n^2} - \frac{a}{2} \bigg| = \, \lim_{n \to \infty }\bigg | \frac{2na_1 + 2(n-1)a_2 + ... +4a_{n-1} + 2a_n - n^2a}{2n^2} \bigg| = \\ \displaystyle \frac{1}{2}\lim_{n \to \infty }\bigg | \frac{(2na_1 - na) + (2(n-1)a_2 - na) + ... + (4a_{n-1} - na) + (2a_n - na)}{n^2} \bigg| = \\ \displaystyle\frac{1}{2}\lim_{n \to \infty} \bigg | \frac{n(2a_1 - a) + n(2\frac{(n-1)}{n}a_2 - a) + ... + n(\frac{4}{n}a_{n-1} - a) + n(\frac{2}{n}a_n - a)}{n^2} \bigg| = \\ \displaystyle\frac{1}{2}\lim_{n \to \infty} \bigg | \frac{(2a_1 - a) + (2\frac{(n-1)}{n}a_2 - a) + ... + (\frac{4}{n}a_{n-1} - a) + (\frac{2}{n}a_n - a)}{n} \bigg|$.
I can continue but I am not sure if this is exactly the right way to go about this.