$\lim_{s\to 1} \sum_{n\ge 0} (a_n-a_{n-1})s^n =\lim_{m\to\infty} \sum_{n=0}^m \frac{a_n}{m+1}$? Suppose $0\le a_n\le 1$ for all $n\ge 0$. Define $a_{-1}=0$. Suppose $\sum\limits_{n=0}^{\infty} a_ns^n$ is finite for all $0<s<1$. But $\sum\limits_{n=0}^{\infty} a_n$ diverges. Is it true that
$$\lim_{s\to 1^{-}} \sum_{n=0}^{\infty} (a_n-a_{n-1})s^n=\lim_{m\to\infty} \frac1{m+1}\sum_{n=0}^{m} a_n$$
whenever right hand side limit exist?
 A: 
The limit on the right is the Cesaro sum of $\sum_{n=0}^{\infty}(a_n-a_{n-1})$, while the limit on the left is the Abel summation of $\sum_{n=0}^{\infty}(a_{n}-a_{n-1})$.
Abel's summation method is stronger than, and consistent with, the Cesaro summation method.

The following is the classical proof:
Denote $b_n=\sum_{k=0}^{n}a_k=\sum_{k=0}^{n}\left(\sum_{i=0}^{k}(a_{i}-a_{i-1})\right)$
Then $\lim_{m\to\infty}\frac{b_m}{m+1}$ is the limit on the right hand side of your equation and assume that it is equal to $L$.
Since $\frac{b_n}{n+1}\to L$ then, for all small $\epsilon>0$ there is an $N$ such that for $n>N$ we have $$L-\epsilon<\frac{b_n}{n+1}<L+\epsilon$$
On the other hand, for $0\leq s<1$,
$$\begin{align}\sum_{n=0}^{\infty}(a_n-a_{n-1})s^n&=\sum_{n=0}^{\infty}b_n(1-s)^2s^n\\&=\frac{\sum_{n=0}^{\infty}b_ns^n}{\sum_{n=0}^{\infty}(n+1)s^n}\\&=\frac{\sum_{n=0}^{N}b_ns^n+\sum_{n=N+1}^{\infty}b_ns^n}{\sum_{n=0}^{N}(n+1)s^n+\sum_{n=N+1}^{\infty}(n+1)s^n}\end{align}$$
If $N(x)$ denotes the numerator and $D(x)$ the denominator above we have
$$N(s)\leq (L+\epsilon)D(s)+\sum_{n=0}^{N}|b_n|s^n$$
and
$$N(s)\geq (L-\epsilon)D(s)-\sum_{n=0}^{N}(n+1)s^n-\sum_{n=0}^{N}|b_n|s^n$$
Then $$(L-\epsilon)-\frac{\sum_{n=0}^{N}ns^n+\sum_{n=0}^{N}|b_n|s^n}{D(s)}\leq\frac{N(s)}{D(s)}\leq (L+\epsilon)+\frac{\sum_{n=0}^{N}|b_n|s^n}{D(s)}$$
Taking $\limsup_{s\to1^-}$ and $\liminf_{s\to1^-}$, and taking into account that $D(s)\to\infty$, we get $$L-\epsilon\leq\liminf_{s\to1^-}\frac{N(s)}{D(s)}\leq\limsup_{s\to1^-}\frac{N(s)}{D(s)}\leq L+\epsilon$$
Since this is for all $\epsilon>0$ we must have $$\liminf_{s\to1^-}\frac{N(s)}{D(s)}=\limsup_{s\to1^-}\frac{N(s)}{D(s)}=\lim_{s\to1^-}\frac{N(s)}{D(s)}=L.$$
A: Remark: I posted this before you edited "whenever right hand side limit exist".
No. The function with
$\dfrac{s}{(1-s)^2}=\sum_{n=1}^\infty n\,s^n$ 
for $s<1$, has $a_n-a_{n-1}=1$ and hence the left hand side is the geometric series $\sum_{n=1}^\infty s^n=\frac{1}{1-s}$. This has a limit of $\frac{1}{2}$ at $s=-1$. 
Meanwhile, the partial sums of all $n$ up to $m$ is a quadratic function and after dividing that by $m-1$ it's still divergent.
