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!

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.