# proof of Abel’s convergence theorem

Suppose that

 $\sum_{n=0}^{\infty}a_{n}=L$

is a convergent series, and set

 $f(r)=\sum_{n=0}^{\infty}a_{n}r^{n}.$

Convergence of the first series implies that $a_{n}\rightarrow 0$, and hence $f(r)$ converges for $|r|<1$. We will show that $f(r)\rightarrow L$ as $r\rightarrow 1^{-}$.

Let

 $s_{N}=a_{0}+\cdots+a_{N},\quad N\in\mathbb{N},$

denote the corresponding partial sums. Our proof relies on the following identity

 $f(r)=\sum_{n}a_{n}r^{n}=(1-r)\sum_{n}s_{n}r^{n}.$ (1)

The above identity obviously works at the level of formal power series. Indeed,

 $\begin{array}[]{crcrcrc}&a_{0}&+&(a_{1}+a_{0})\,r&+&(a_{2}+a_{1}+a_{0})\,r^{2}% &+\,\cdots\\ -\,(&&&a_{0}\,r&+&(a_{1}+a_{0})\,r^{2}&+\,\cdots)\\ =&a_{0}&+&a_{1}\,r&+&a_{2}\,r^{2}&+\,\cdots\end{array}$

Since the partial sums $s_{n}$ converge to $L$, they are bounded, and hence $\sum_{n}s_{n}r^{n}$ converges for $|r|<1$. Hence for $|r|<1$, identity (1) is also a genuine functional equality.

Let $\epsilon>0$ be given. Choose an $N$ sufficiently large so that all partial sums, $s_{n}$ with $n>N$, satisfy $|s_{n}-L|\leq\epsilon$. Then, for all $r$ such that $0, one obtains

 $\left|\sum_{n=N+1}^{\infty}(s_{n}-L)r^{n}\right|\leq\epsilon\,\frac{r^{N+1}}{1% -r}\,.$

Note that

 $f(r)-L=(1-r)\sum_{n=0}^{N}(s_{n}-L)r^{n}+(1-r)\sum_{n=N+1}^{\infty}(s_{n}-L)r^% {n}.$

As $r\rightarrow 1^{-}$, the first term tends to $0$. The absolute value of the second term is estimated by $\epsilon\,r^{N+1}\leq\epsilon$. Hence,

 $\limsup_{r\rightarrow 1^{-}}|f(r)-L|\leq\epsilon.$

Since $\epsilon>0$ was arbitrary, it follows that $f(r)\rightarrow L$ as $r\rightarrow 1^{-}$. QED

Title proof of Abel’s convergence theorem ProofOfAbelsConvergenceTheorem 2013-03-22 13:07:39 2013-03-22 13:07:39 rmilson (146) rmilson (146) 9 rmilson (146) Proof msc 40G10 ProofOfAbelsLimitTheorem