Weierstrass M-Test

Theorem

Suppose:

• $f_n$ is a sequence of functions defined on a domain $D$;

• $\displaystyle \sup_{x\in D} |f_n (x)| \leq M_n$ for each integer $n$ and some constants $M_n$;

• $\displaystyle \sum_{i=1}^{\infty} M_i < \infty$.

Then $\displaystyle \sum_{i=1}^{\infty} f_i$ converges uniformly on $D$.

Proof

Let $\displaystyle S_n = \sum_{i=1}^n f_i$, and let $\displaystyle f = \lim_{n\to \infty}S_n$.

To show the partial sums converge uniformly to $f$, we must show that $\displaystyle \lim_{n\to\infty}\sup_{x\in D} |f - S_n| = 0$.

But:

$\displaystyle \sup_{x\in D} |f - S_{n}| = \sup_{x\in D} |(f_{1} + f_{2} + ...) - (f_{1} + f_{2} + ... + f_{n})| = \sup_{x\in D} |f_{n+1} + f_{n+2} + \ldots|$

By the Triangle Inequality, this value is less than or equal to $\displaystyle \sum_{i=n+1}^{\infty}\sup_{x\in D}|f_{i}(x)| \leq \sum_{i=n+1}^{\infty}M_{i}$.

But since $\displaystyle 0 \leq \sum_{i=1}^{\infty}M_{n} < \infty$, and a convergent series has tails that converge to zero, it follows that:

$\displaystyle 0 \leq \lim_{n\to\infty}\sum_{i=n+1}^{\infty}\sup_{x\in D}|f_{i}(x)| \leq \lim_{n\to\infty}\sum_{i=n+1}^{\infty}M_{i} = 0$

So $\displaystyle \lim_{n\to\infty}\sup_{x\in D}|f - S_{n}| = 0$.

Hence the series converges uniformly on the domain.

$\blacksquare$

Source of Name

This entry was named for Karl Theodor Wilhelm Weierstrass.