📂Complex Anaylsis

Weierstrass M-Test

Theorem ¹

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

Explanation

The name M-test comes from the sequence $M_{n}$. If we can bring in a suitable $M_{n}$ that is already known to converge and set up an inequality with the absolute value of the function, then we can show not merely convergence but absolute convergence and uniform convergence at the same time, which makes it a useful theorem. Above all, once the inequality has been established, it is convenient because we only need to think about a sequence of real numbers.

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 hypothesis of the theorem, $\displaystyle \sum_{n=1}^{\infty} |f_{n}(z)|$ converges, and by the definition of absolute convergence we can say that it converges absolutely.

Next, for uniform convergence we use the Cauchy criterion.

Let $R_{k}(z)$ be the sum of the terms following the $k$th term in $\displaystyle \sum_{n=1}^{\infty} f_{n}(z)$, and let $R_{k}^{ \ast }$ be the sum of the terms following the $k$th term in $\displaystyle \sum_{n=1}^{\infty} M_{n}$. 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: $\displaystyle \sum _{ n=1 }^{ \infty }{ { a }_{ n }}$ converges if and only if $\displaystyle \lim_{n \to \infty} \sum _{ k=n }^{ n+m }{ { a }_{ k }}=0$.

By the Cauchy criterion, $\displaystyle \lim_{k \to \infty} R_{k}^{ \ast } = 0$, so $\displaystyle \lim_{k \to \infty} |R_{k}(z)| = 0$, that is, $\displaystyle \lim_{k \to \infty} R_{k}(z) = 0$. Since the above argument can be applied for every $z \in A$, $\displaystyle \sum_{n=1}^{\infty} f_{n}(z)$ converges uniformly on $A$.

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