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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?

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    $\begingroup$ The right-hand side is the Cesaro sum of $\sum_{n=0}^{\infty}(a_{n}-a_{n-1})$, while the left-hand side is the Abel's sum of the same series. It is known that Abel's summation method is stronger than (and consistent with) Cesaro summation. This means that whenever Cesaro summation gives a number so does Abel's and it gives the same result. So, the answer is yes. $\endgroup$ – Carol Feb 9 '15 at 16:32
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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 $$\liminf_{s\to1^-}\frac{N(s)}{D(s)}=\limsup_{s\to1^-}\frac{N(s)}{D(s)}=\lim_{s\to1^-}\frac{N(s)}{D(s)}=L.$$

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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.

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  • $\begingroup$ It should be $s \to 1^{-}$, not $s = -1$ $\endgroup$ – graydad Feb 9 '15 at 15:55
  • $\begingroup$ @graydad: The limit to 1 from the left? Okay, I see. $\endgroup$ – Nikolaj-K Feb 9 '15 at 16:00
  • $\begingroup$ sorry i forgot to add that $a_n\le 1$. actually those are probabilities of some events. can you still find a counterexample? $\endgroup$ – Sayan Feb 9 '15 at 16:00
  • $\begingroup$ Yup. I found that $$\lim_{s\to 1^{-}} \sum_{n=0}^{\infty} (a_n-a_{n-1})s^n = \lim_{s\to 1^{-}} \sum_{n=0}^{\infty} a_ns^n(1-s)$$ but am stuck at that point. $\endgroup$ – graydad Feb 9 '15 at 16:01

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