Abel's Summability

Abel's Theorem

Suppose that the power series $\sum _{n=0}^{\mathrm{\infty }}{a}_n{x}^n$ converges to $f(x)$ for $|x|<1$ and that $\sum _{n=0}^{\mathrm{\infty }}{a}_n$ converges to $A$. Then the power series converges uniformly on $[0,1]$ and

A series $\sum _{n=0}^{\mathrm{\infty }}{a}_n$ is said to be Abel Summable to $L$ if the power series,

converges in $[0,1)$ and $L=\underset{x\to 1^{-}}{lim}f(x)$. It can be expressed as $\sum _{n=0}^{\mathrm{\infty }}{a}_n=L(\text{Abel})$

If $\sum _{k=1}^{\mathrm{\infty }}{a}_k=L$ in the usual sense then $\sum _{k=1}^{\mathrm{\infty }}{a}_k=L(\text{Abel})$

Tauber's Theorem

Suppose that the power series $\sum _{n=0}^{\mathrm{\infty }}{a}_n{x}^n$ converges to $f(x)$ for $|x|<1$ and $limn{a}_n=0$, or $a_n=o(1/n)$. If $\underset{x\to 1^{-}}{lim}f(x)=A$, then the series $\sum _{n=0}^{\mathrm{\infty }}{a}_n$ converges to $A$.

Littlewood's Tauberian Theorem

Suppose that the power series $\sum _{n=0}^{\mathrm{\infty }}{a}_n{x}^n$ converges to $f(x)$ for $|x|<1$ and $a_n=O(1/n)$. If $\underset{x\to 1^{-}}{lim}f(x)=A$, then the series $\sum _{n=0}^{\mathrm{\infty }}{a}_n$ converges to $A$.

We have that for series $$\text{summable} \implies \text{Cesàro summable} \implies \text{Abel summable}$$