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\newtheorem*{theorem}{Theorem}

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\title{Abel's Theorem}

\author{Math 166: Calculus II}

\date{April 16, 1999}

\begin{document}

\maketitle

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\begin{theorem}[Abel]

Suppose $\sum_{n=0}^\infty a_n$ converges. Then

  \[
  f(x) = \sum_{n=0}^\infty a_n x^n
  \]
converges absolutely for $|x|<1$ and
  \[
  \lim_{x\to 1^-} f(x) = \sum_{n=0}^\infty a_n
  \]

\end{theorem}

\begin{proof}
Since $a_n \to 0$, there is a positive integer $N$ such that $|a_n| <
1$ for all $n \ge N$. Then $|a_nx^n| = |a_n| |x|^n < |x|^n$ for all $n\ge
N$. The series $\sum_{n=0}^\infty |x|^n$ converges for $|x|<1$, so the
series $\sum_{n=0}^\infty |a_nx^n|$ converges by the Comparison Theorem.

To prove the statement about the limit, we must show that given $\epsilon
> 0$ there is a $\delta > 0$ such that if $1-\delta < x < 1$ then $|f(x)
- s| < \epsilon$ where $s = \sum_{n=1}^\infty a_n$. The usual strategy
is to try to write $f(x) - s = (1-x)\cdot(\mbox{something})$ so that as
$x\to1$, $f(x)\to s$. Abel found a clever way to do this with series:
Let $s_k = \sum_{n=0}^k a_n$ and define $s_{-1} = 0$, so that $a_n =
s_n - s_{n-1}$ for all $n \ge 0$. Then,
  \begin{eqnarray*}
  \sum_{n=0}^k a_n x^n &=& \sum_{n=0}^k (s_n - s_{n-1})x^n
     = \sum_{n=0}^k s_nx^n - \sum_{n=1}^k s_{n-1}x^n\\
     &=& \sum_{n=0}^k s_nx^n - \sum_{n=0}^{k-1} s_{n}x^{n+1}\quad
         \mbox{[shift index]}\\
     &=& s_k x^k + \sum_{n=0}^{k-1} s_nx^n(1 - x)\\
  \end{eqnarray*}
Now assume $|x|<1$ and let $k\to \infty$. Since $s_kx^k\to s\cdot0 = 0$, we
find
  \[
  f(x) = \sum_{n=0}^\infty a_nx^n = (1-x)\sum_{n=0}^\infty s_n x^n
  \]
The formula for a geometric series implies that $1 =
(1-x)\sum_{n=0}^\infty x^n$,\newline so $s = (1-x)\sum_{n=0}^\infty sx^n$
and
  \[
  f(x) - s = (1-x)\sum_{n=0}^\infty s_n x^n - (1-x)\sum_{n=0}^\infty sx^n
           = (1-x)\sum_{n=0}^\infty (s_n  - s)x^n
  \]
Given $\epsilon > 0$, there is a positive integer $N$ such that $|s_n - s|
< \frac{\epsilon}{2}$ whenever $n\ge N$. Therefore, assuming $0<x<1$,
  \begin{eqnarray*}
  |f(x)-s| &=& (1-x) \Big|\sum_{n=0}^\infty (s_n-s)x^n\Big|
    \le (1-x)\sum_{n=0}^\infty|s_n-s|x^n\\
    &=& (1-x)\sum_{n=0}^N|s_n-s|x^n+(1-x)\sum_{n=N}^\infty|s_n-s|x^n\\
    &<& (1-x)\sum_{n=0}^N|s_n-s|+(1-x)\sum_{n=N}^\infty\frac{\epsilon}{2}x^n\\
    &<& (1-x) K + \frac{\epsilon}{2}\\
  \end{eqnarray*}
where $K = \sum_{n=0}^N |s_n-s|$. The last inequality follows by adding
the geometric series: $(1-x) \sum_{n=N}^\infty \frac{\epsilon}{2} x^n =
\frac{\epsilon}{2} x^N < \frac{\epsilon}{2}$.  We can make $(1-x)K$ small
too: $(1-x)K < \frac{\epsilon}{2}$ if and only if $1 - \frac{\epsilon}{2K}
< x$. Therefore, if we take $\delta = \frac{\epsilon}{2K}$, then
$1-\delta<x<1$ implies that
  \[ |f(x) - s| < (1-x)K + \frac{\epsilon}{2} < \frac{\epsilon}{2}
  +\frac{\epsilon}{2} =  \epsilon \]
which shows $\lim_{x\to1^-}f(x) = s$.

\end{proof}

\end{document}
