Theorem:6481641: Difference between revisions

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theorem

Named after: Niels Henrik Abel

This page was built for theorem: Abel's theorem

Theorem

Suppose that $\sum_{n = 0}^{\infty} a_{n} x^{n}$ has a radius of convergence $r$ and that $\sum_{n = 0}^{\infty} a_{n} r^{n}$ is convergent. Then $\lim_{x \to r^{-}} \sum_{n = 0}^{\infty} a_{n} x^{n} = \sum_{n = 0}^{\infty} a_{n} r^{n} .$

Proof

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

Let $s_{N} = a_{0} + a_{1} + \dots + a_{N}, N \in ℕ$ denote the corresponding partial sums. Our proof relies on the following identity: $f (r) = \sum_{n = 0}^{\infty} a_{n} r^{n} = (1 - r) \sum_{n = 0}^{\infty} s_{n} r^{n} .$ (1)

The above identity obviously works at the level of formal power series. Indeed, $a_{0} + (a_{1} + a_{0}) r + (a_{2} + a_{1} + a_{0}) r^{2} + \dots - (a_{0} r + (a_{1} + a_{0}) r^{2} + \dots) = a_{0} + a_{1} r + a_{2} r^{2} + \dots$

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

Let $ϵ > 0$ be given. Choose an $N$ sufficiently large so that all partial sums $s_{n}$ with $n > N$ satisfy $| s_{n} - L | \leq ϵ$ Then, for all $r$ such that $0 < r < 1$ , one obtains $| \sum_{n = N + 1}^{\infty} (s_{n} - L) r^{n} | \leq \frac{ϵ r^{N + 1}}{1 - r}$

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 $r \to 1^{-}$ , the first term tends to 0. The absolute value of the second term is estimated by $ϵ r^{N + 1} \leq ϵ$ Hence, $\underset{r \to 1^{-}}{lim sup} | f (r) - L | \leq ϵ$

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

Lean Prover

Abel's Theorem

Sources

Abel’s limit theorem in Planetmath

Proof of Abel's Convergence Theorem in Planetmath

@@ Line 1: / Line 1: @@
 {{Theorem}}
-If a real or complex power series for a function has radius of convergence 1 and the series is only known to converge conditionally at 1, Abel's limit theorem gives the value at 1 as the limit of the function at 1 from the left. "Left" for complex numbers means within a fixed cone opening to the left with angle less than Pi.
+==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>
-Test visual edit.
+==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>.
+==Lean Prover==
+[https://leanprover-community.github.io/mathlib4_docs/Mathlib/Analysis/Complex/AbelLimit.html#Complex.tendsto_tsum_powerSeries_nhdsWithin_stolzCone Abel's Theorem]
+==Sources==
+[https://planetmath.org/abelslimittheorem Abel’s limit theorem in Planetmath]
+[https://planetmath.org/proofofabelsconvergencetheorem Proof of Abel's Convergence Theorem in Planetmath]