Let $\{s_n\} \subset \mathbb{C}$. Define $$\sigma_n := \frac{s_0 + \cdots + s_n}{n+1}.$$ (e) Let $a_n := s_n - s_{n-1}$ for $n \geq 1$. Assume further that $|na_n| \leq M$ for all $n$, and $\sigma_n \to \sigma$. Prove that $\lim s_n = \sigma$. Hint: for $m < n$, we have that $$s_n - \sigma_n = \frac{m+1}{n-m}(\sigma_n - \sigma_m) + \frac{1}{n-m} \sum_{i=m+1}^n (s_n - s_i).$$ For $m+1 \leq i \leq n$, we have that $$ |s_n - s_i| \leq \frac{(n-i)M}{i+1} \leq \frac{(n-m-1)M}{m+2}. $$ Fix $\epsilon > 0$ and for each $n$, we find some $m$ such that $$m \leq \frac{n-\epsilon}{1+\epsilon} < m+1.$$ Then it follows that $\frac{m+1}{n-m} \leq \epsilon^{-1}$ and $|s_n - s_i| < M\epsilon$. Hence $\limsup_{n\to \infty} |s_n - \sigma| \leq M\epsilon$.
Following the hint, I think our goal is to show that $\limsup_{n \to \infty} |s_n - \sigma_n| = 0$. I could prove the first equality concerning $s_n - \sigma_n$, and get the second inequality regarding $|s_n-s_i|$. Then plug in, we have that $$ |s_n - \sigma_n| \leq \underbrace{\frac{m+1}{n-m}}_{(\text{I})}(\sigma_n - \sigma_m) + \underbrace{\frac{n-m-1}{m+2}}_{(\text{II})}M. $$ Since we have the convergence of $\sigma_n$, by Cauchyness, we could control $(\sigma_n - \sigma_m)$. Therefore at this stage, we would like to control (I) and (II), so that (I) is somehow "bounded", and make (II) as small as we want.
The hint suggests that we should find corresponding $m$, then if make $n, m$ large. so that $\sigma_n - \sigma_m \leq \epsilon^2$, then we are done.
My Question: Without the hint, I think it is natural to come up with the first equality and the second inequality. But how to come up with such $m$, to control (I) and (II)? More precisely, why would we choose the integer $m$ to satisfy that
$$m \leq \frac{n-\epsilon}{1+\epsilon} < m+1?$$
I am afraid that my description is somewhat vague, so please let me know if I could make it more clear. Thank you!