View source for Theorem:6481641

{{Theorem}}

==Theorem==

Suppose that <math>\sum_{n=0}^{\infty} a_n x^n</math> has a radius of convergence <math>r</math> and that <math>\sum_{n=0}^{\infty} a_n r^n</math> is convergent. Then
<math display="block"> 
\lim_{x \to r^-} \sum_{n=0}^{\infty} a_n x^n = \sum_{n=0}^{\infty} a_n r^n. 
</math>

==Proof==
Suppose that
<math display="block">
  \sum_{n=0}^{\infty} a_n = L
</math>
is a convergent series, and define
<math display="block">
  f(r) = \sum_{n=0}^{\infty} a_n r^n
</math>
The convergence of the first series implies that <math>a_n \to 0</math>, and hence <math>f(r)</math> converges for <math>|r| < 1</math>. We will show that <math>f(r) \to L</math> as <math>r \to 1^-</math>.

Let
<math display="block">
  s_N = a_0 + a_1 + \cdots + a_N, \quad N \in \mathbb{N}
</math>
denote the corresponding partial sums. Our proof relies on the following identity:
<math display="block">
  f(r) = \sum_{n=0}^{\infty} a_n r^n = (1 - r) \sum_{n=0}^{\infty} s_n r^n.
</math>
(1)

The above identity obviously works at the level of formal power series. Indeed, 
<math display="block">
  a_0 + (a_1 + a_0)r + (a_2 + a_1 + a_0)r^2 + \cdots - \left( a_0 r + (a_1 + a_0) r^2 + \cdots \right) = a_0 + a_1 r + a_2 r^2 + \cdots
</math>

Since the partial sums <math>s_n</math> converge to <math>L</math>, they are bounded, and hence
<math display="block">
  \sum_{n=0}^{\infty} s_n r^n
</math>
converges for <math>|r| < 1</math>. Hence, for <math>|r| < 1</math>, identity (1) is also a genuine functional equality.

Let <math> \epsilon > 0</math> be given. Choose an <math xmlns="http://www.w3.org/1998/Math/MathML">N</math> sufficiently large so that all partial sums <math xmlns="http://www.w3.org/1998/Math/MathML">s_n</math> with <math xmlns="http://www.w3.org/1998/Math/MathML">n > N</math> satisfy
<math display="block">
  |s_n - L| \leq \epsilon
</math>
Then, for all <math xmlns="http://www.w3.org/1998/Math/MathML">r</math> such that <math xmlns="http://www.w3.org/1998/Math/MathML">0 < r < 1</math>, one obtains
<math display="block">
  \left| \sum_{n=N+1}^{\infty} (s_n - L) r^n \right| \leq \frac{\epsilon r^{N+1}}{1 - r}
</math>

Note that
<math display="block">
  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
</math>

As <math xmlns="http://www.w3.org/1998/Math/MathML">r \to 1^-</math>, the first term tends to 0. The absolute value of the second term is estimated by
<math display="block">
  \epsilon r^{N+1} \leq \epsilon
</math>
Hence,
<math display="block">
  \limsup_{r \to 1^-} |f(r) - L| \leq \epsilon
</math>

Since <math xmlns="http://www.w3.org/1998/Math/MathML"> \epsilon > 0</math> was arbitrary, it follows that <math xmlns="http://www.w3.org/1998/Math/MathML">f(r) \to L</math> as <math xmlns="http://www.w3.org/1998/Math/MathML">r \to 1^-</math>.

==Sources==
[https://planetmath.org/abelslimittheorem Abel’s limit theorem in Mathplanet]
[https://planetmath.org/proofofabelsconvergencetheorem Proof of Abel's Convergence Theorem in Mathplanet]