📂Complex Anaylsis

Weierstrass M-test

Theorem ¹

For functions $f_{n}$ and $z \in A$, if there exists a sequence of positive numbers $M_{n}$ that satisfies $|f_{n}(z)| \le M_{n}$ and $\displaystyle \sum_{n=1}^{\infty} M_{n}$ converges, then $\displaystyle \sum_{n=1}^{\infty} f_{n}$ absolutely converges and uniformly converges in $A$.

Description

The M-test is named after the sequence $M_{n}$. It is a useful theorem that allows us to show both absolute and uniform convergence at once if we can bring in the already known converging sequence $M_{n}$, establish the inequality with the absolute value of the function. Most conveniently, after establishing the inequality, only the sequence of real numbers needs to be considered.

Proof

Absolute convergence can be shown very easily.

Alternating Series Test: If $b_n \downarrow 0$, then $\displaystyle \sum _{ n=1 }^{ \infty }{ (-1)^{n} {b}_{n}}$ converges.

By the comparison test and the assumption of the theorem, $\displaystyle \sum_{n=1}^{\infty} |f_{n}(z)|$ converges, and it can be said to converge absolutely according to the definition of absolute convergence.

Next, uniform convergence is proved using the Cauchy criterion.

If we call the sum of the terms following the $k$th in $\displaystyle \sum_{n=1}^{\infty} f_{n}(z)$ as $R_{k}(z)$, and the sum of the terms following the $k$th in $\displaystyle \sum_{n=1}^{\infty} M_{n}$ as $R_{k}^{ \ast }$, then the following holds. $$|R_{k}(z)| = \left| \sum_{n=k+1}^{\infty} f_{n}(z) \right| \le \sum_{n=k+1}^{\infty} |f_{n}(z)| \le \sum_{n=k+1}^{\infty} M_{n} = R_{k}^{ \ast }$$

Cauchy Criterion: The convergence of $\displaystyle \sum _{ n=1 }^{ \infty }{ { a }_{ n }}$ is equivalent to $\displaystyle \lim_{n \to \infty} \sum _{ k=n }^{ n+m }{ { a }_{ k }}=0$.

By the Cauchy criterion, since $\displaystyle \lim_{k \to \infty} R_{k}^{ \ast } = 0$, it follows $\displaystyle \lim_{k \to \infty} |R_{k}(z)| = 0$, i.e., $\displaystyle \lim_{k \to \infty} R_{k}(z) = 0$. Since the above discussion can be applied to all $z \in A$, $\displaystyle \sum_{n=1}^{\infty} f_{n}(z)$ uniformly converges in $A$.

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