The question posed is as follows:
For $\{s_n\}$ a sequence of complex numbers, $\sigma_n = \frac{s_0 + s_1 + ... + s_n}{n+1}$, $a_n = s_n - s_{n-1}$, $| n a_n | \leq M < \infty$, $ \forall n \in \mathbb{N}$, and $\lim \sigma_n = \sigma$:
Prove that $\lim s_n = \sigma$ by completing the following outline.
If $m < n$, then $s_n - \sigma_n = \frac{m+1}{n-m} (\sigma_n - \sigma_m) + \frac{1}{n-m} \sum_{i=m+1}^{n} (s_n - s_i)$.
For $i \in \{m+1, ... , n\}$,
$| s_n - s_i | \leq \frac{(n-i) M}{i+1} \leq \frac{(n-m-1)M}{m+2}$.
And fixing any $\epsilon > 0$, we can choose an integer $m$ for each $n$ such that the following is satisfied,
$m \leq \frac{n-\epsilon}{1+\epsilon} < m+1$ so that $\frac{(m+1)}{(n-m)} \leq \frac{1}{\epsilon}$ and $|s_n - s_i| < M \epsilon$.
The text then states
"Hence, $\displaystyle{\limsup_{n \to \infty}} |s_n - \sigma_n| \leq M \epsilon$, since $\epsilon$ was arbitrary, $\lim s_n = \sigma$. "
My question
I follow everything up to the last line of the text in quotes.
Using every step of the outline, I arrive at:
$|s_n - \sigma_n| < \frac{1}{\epsilon}|\sigma_n - \sigma_m| + M \epsilon$.
I guess I'm not seeing how this would then imply that $|s_n - \sigma_n| < M \epsilon$. Unless it could be shown that some Cauchy subsequence for $\sigma_n, \sigma_m$, that satisfies that pair of m,n chosen above converges faster than $\epsilon$? Otherwise it would simply blow up. But I'm not sure how I'm supposed to arrive there or get rid of the $\frac{1}{\epsilon}$ factor in the bound.
Take the $\limsup$ on both sides directly without trying to prove what you wanted to prove: $$\limsup_{n \to \infty} \left|s_n - \sigma_n\right| \leq \limsup_{n \to \infty}\left(\frac{1}{\varepsilon}\left|\sigma_n - \sigma_{m(n)}\right| + M \varepsilon\right) = \frac{1}{\varepsilon} \limsup_{n \to \infty}\left|\sigma_n - \sigma_{m(n)}\right| + M\varepsilon$$ Now, we have $m(n) \leq n$ and $m(n) \to \infty$ when $n \to \infty$, and by convergence of $(\sigma_n)_n$ there exists for all $\delta > 0$ some $N_\delta \in \mathbb{N}$ such that for all $n,n' \geq N_{\delta}$, $|\sigma_n - \sigma_{n'}| \leq \delta$.
Combining these facts, we get the existence of $N_\delta'$ such that, for all $n \geq N_{\delta}'$, we have $n \geq m(n) \geq N_\delta$, and hence such that: $$\forall n \geq N_\delta',\quad \left|\sigma_n - \sigma_{m(n)}\right| \leq \delta$$ which is exactly saying that $\left|\sigma_n - \sigma_{m(n)}\right| \to 0$ when $n \to \infty$.
Therefore: $$\limsup_{n \to \infty} |s_n - \sigma_n| \leq \frac{1}{\varepsilon} \limsup_{n \to \infty}\left|\sigma_n - \sigma_{m(n)}\right| + M\varepsilon = M\varepsilon$$ and we are done.