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 $a_{n}\\rightarrow 0$ , andhence $f(r)$ 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 thefollowing 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 powerseries. 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, andhence $\\sum_{n}s_{n}r^{n}$ 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 thatall 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 thesecond term is estimated by $\\epsilon\\,r^{N+1}\\leq\\epsilon$ . Hence,

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

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