A theorem about Cesàro mean, related to Stolz-Cesàro theorem Original Title: Tauberian theorems and Cesàro sum
Theorem (Landau-Hardy, From Rudin's Principle of Mathematical Analysis Exercise 3.14)
$\newcommand\abs[1]{\left\lvert#1\right\rvert}$
If $\{s_n\}$ is a complex sequence, define its arithmetic means $\sigma_n$ by
$$\sigma_n=\frac{s_0+s_1+\dotsb+s_n}{n+1}\qquad(n=0,1,2,\dotsc)$$
Put $a_n=s_n-s_{n-1}$ for $n\ge1$.
Assume $M<+\infty$ and $\abs{na_n}\le M$ for all $n$, and $\lim_{n\to\infty}\sigma_n=\sigma$, then $\lim_{n\to\infty}s_n=\sigma$.
The outline of the proof
If $m<n$, then
$$s_n-\sigma_n=\frac{m+1}{n-m}(\sigma_n-\sigma_m)+\frac1{n-m}\sum_{k=m+1}^n(s_n-s_k)\tag{*}$$
Notice that $\abs{s_n-s_k}\le(n-m-1)M\,/\,(m+2)$, fix $\epsilon>0$ and associate with each $n$ the integer $m$ that satisfies
$$m\le\frac{n-\epsilon}{1+\epsilon}<m+1$$
Then $(m+1)\,/\,(n-m)\le1/\epsilon$ and $\abs{s_n-s_k}<M\epsilon$. Hence
$$\limsup_{n\to\infty}\,\abs{s_n-\sigma}\le M\epsilon$$
Questions and thoughts
It seems that the equation (*) comes out strangely. I wonder how to discover such kind of strange identities. So is there any observation, even deeper, to look through that equation?
Thanks!
 A: It is not clear exactly what is being asked, but here is my take on what is going on in the proof and how it applies to $(\ast)$.
The idea of the proof is to use that $\sigma_n$ is Cauchy and that $s_n-s_{n-1}$ is small to estimate $s_n-\sigma_n$.
Start with a simple equation which localizes the average of $s_k$ for large $k$
$$
\sum_{k=m+1}^ns_k=\color{#C00000}{\sum_{k=0}^ns_k}-\color{#00A000}{\sum_{k=0}^ms_k}\tag{1}
$$
and rewrite the red and green sums using $\sigma_n$
$$
\sum_{k=m+1}^ns_k=\color{#C00000}{(n+1)\sigma_n}-\color{#00A000}{(m+1)\sigma_m}\tag{2}
$$
This writes things nicely as $(n-m)$ sigmas, so we subtract from a like number of $s_n$ to get closer to the goal of $s_n-\sigma_n$.
Subtract both sides of $(2)$ from $\displaystyle\sum_{k=m+1}^ns_n=(n-m)s_n$ to exploit the small size of $s_n-s_{n-1}$
$$
\begin{align}
\sum_{k=m+1}^n(s_n-s_k)
&=(n-m)s_n-\Big[(n+1)\sigma_n-(m+1)\sigma_m\Big]\\
&=(n-m)s_n-\Big[(n-m)\sigma_n+(m+1)\sigma_n-(m+1)\sigma_m\Big]\\[8pt]
&=(n-m)(s_n-\sigma_n)-(m+1)(\sigma_n-\sigma_m)\tag{3}
\end{align}
$$
This gives the desired quantity, $s_n-\sigma_n$, as a sum of controllable terms: $s_n-s_k$ and $\sigma_n-\sigma_m$.
Add $(m+1)(\sigma_n-\sigma_m)$ to both sides to isolate $s_n-\sigma_n$
$$
(m+1)(\sigma_n-\sigma_m)+\sum_{k=m+1}^n(s_n-s_k)=(n-m)(s_n-\sigma_n)
$$
Divide both sides by $n-m$ to get $(\ast)$
$$
\frac{m+1}{n-m}(\sigma_n-\sigma_m)+\frac1{n-m}\sum_{k=m+1}^n(s_n-s_k)=(s_n-\sigma_n)\tag{$\ast$}
$$
For large $m$ and $n$, $\sigma_n-\sigma_m$ is small since $\sigma_n$ is Cauchy and $\displaystyle\sum_{k=m+1}^n(s_n-s_k)$ is small because $s_n-s_{n-1}$ is small. This is the general idea; the details are in the outline of the proof.
