Help with Baby Rudin Excercise 3.17. Partial converse to the fact that convergence impiies Cesaro summability. Warning: Pretty involved problem. If $\{s_n\}$ is a complex sequence, define its arithmetic mean $\sigma_n$, by
$$\sigma_n=\frac{s_0+s_1+\cdots+s_n}{n+1}.$$
Put $a_n=s_n-s_{n-1}$ for $n\ge 1$.
Assume $M<\infty$, $|na_n|\le M$ for all $n$, and $\lim\sigma_n=\sigma$. Prove that $\lim s_n=\sigma$, by completing to following outline:
If $m<n$, then
$$s_n-\sigma_n={m+1\over n-m}(\sigma_n-\sigma_m)+\frac1{n-m}\sum_{i=m+1}^n(s_n-s_i).\tag{1}\label{1}$$
For the $k$ in the last term, show 
$$|s_n-s_k|\le{n-m-1\over m+2}M.\tag{2}\label{2}$$
Fix $\epsilon>0$ and associate with each $n$ the integer $m$ that satisfies
$$m\le{n-\epsilon\over 1+\epsilon}<m+1.$$
Then 
$$(m+1)/(n-m)\le\frac1\epsilon \text{ and } |s_n-s_k|\le M\epsilon.\tag{3}\label{3}$$
Hence
$$\limsup_{n\to\infty}|s_n-\sigma|\le M\epsilon.\tag{4}\label{4}$$
I've worked through $\eqref{1}-\eqref{3}$, but having trouble with $\eqref{4}$.
My attempt:
Choose $m$ and $n$ so that $|\sigma-\sigma_m|<\epsilon$, then
$$\begin{align}
\limsup_{n\to\infty}|s_n-\sigma| & \le\limsup_{n\to\infty}|s_n-\sigma_n|+\limsup_{n\to\infty}|\sigma_n-\sigma|\\
& =\limsup_{n\to\infty}|s_n-\sigma_n|\\
& \le{m+1\over n-m}\limsup_{n\to\infty}|(\sigma_n-\sigma_m)|\\
&\quad\quad+\frac1{n-m}\limsup_{n\to\infty}\sum_{k=m+1}^n|s_n-s_k|\\
& \le\frac1\epsilon|\sigma-\sigma_m|+M\epsilon\\
&\color{red}{\le 1+M\epsilon}.
\end{align}$$
I don't know how to do this without getting the $1$ term.

I agree with most of your work. A couple of notes:


*

*"Choose $m$ and $n$ so that $|\sigma-\sigma_m|<\epsilon$" doesn't really make sense since $m$ is predetermined by your choice of $n$ and $\epsilon$.

*All of the instances of $n$ and $m$ should still be contained within the $\limsup$'s.


So with these small alterations you can still get 
$$\limsup_{n\to\infty}|s_n-\sigma|\leq\frac{1}{\epsilon}\limsup_{n\to\infty}|\sigma_n-\sigma_m|+M\epsilon.$$
But since $\{\sigma_n\}$ is convergent and $\frac{n-\epsilon}{1+\epsilon}<m+1$ implies $m\to\infty$ as $n\to\infty$, we have
$$\limsup_{n\to\infty}|\sigma_n-\sigma_m|=0$$
and the desired result follows.
