Series of Functions

Section 7.2 Series of Functions

Definition 7.16.

Let

(f_{n})_{n = 1}^{\infty} \subseteq B (X)

for a set

X .

We'll say

\sum_{n = 1}^{\infty} f_{n}

converges uniformly (pointwise) if

(\sum_{n = 1}^{N} f_{n})_{N = 1}^{\infty} \subseteq B (X)

converges uniformly (pointwise).

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Theorem 7.17. Weierstrass M-test.

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(f_{n})_{n = 1}^{\infty} \subseteq B (X)

for a set

X

and

\sum_{n = 1}^{\infty} {| | f_{n} | |}_{\infty}

converges, then

\sum_{n = 1}^{\infty} f_{n}

converges uniformly (or, equivalently, converges wrt

d_{\infty}

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Remark 7.18.

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This is a slightly non-standard version of the theorem. Combining this version with the comparison test gives the usual statement: If

(f_{n})_{n = 1}^{\infty} \subseteq B (X)

and

(M_{n})_{n = 1}^{\infty} \subseteq [0, \infty)

so that

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for all $x \in R, n \geq 1$ have $| f_{n} (x) | \leq M_{n}$
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$\sum_{n = 1}^{\infty} M_{n} < \infty$

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then

\sum_{n = 1}^{\infty} f_{n}

converges uniformly.

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Example 7.19.

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The series

\sum_{n = 1}^{\infty} \frac{1}{n^{2} + x^{2}}

converges uniformly over

x \in R .

Solution.

Let

M_{n} = \frac{1}{n^{2}} .

Then

| \frac{1}{n^{2} + x^{2}} | \leq \frac{1}{n^{2}}

for all

x \in R

and

\sum_{n = 1}^{\infty} \frac{1}{n^{2}} = \frac{π^{2}}{6} < \infty,

so by the Weierstass

M

-test,

\sum_{n = 1}^{\infty} \frac{1}{n^{2} + x^{2}}

converges uniformly over

x \in R .

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Exploration 7.20.

For any

r > 0,

the series

\sum_{n = 0}^{\infty} \frac{z^{n}}{n!}

converges uniformly on

{z \in C : | z | \leq r},

but

\sum_{n = 0}^{\infty} \frac{z^{n}}{n!}

does not converge uniformly on

C .

Mathematical Analysis