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 $0<r<1$ , 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
Canonical name	ProofOfAbelsConvergenceTheorem
Date of creation	2013-03-22 13:07:39
Last modified on	2013-03-22 13:07:39
Owner	rmilson (146)
Last modified by	rmilson (146)
Numerical id	9
Author	rmilson (146)
Entry type	Proof
Classification	msc 40G10
Related topic	ProofOfAbelsLimitTheorem