A theorem about Cesàro mean, related to Stolz-Cesàro theorem

Question

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!

Answer 1

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.