For a series of functions $\sum f_k$, uniform convergence of the partial sums is hard to check from the definition. The M-test reduces it to a single series of numbers, using comparison and the sup norm.
🔑 Key idea
Weierstrass M-test. If there are numbers $M_k$ with $|f_k(x)| \le M_k$ for every $x \in S$ and $\sum M_k$ converges, then $\sum f_k$ converges uniformly (and absolutely) on $S$. The natural choice is $M_k = \|f_k\| = \sup_S |f_k|$.
💡 Why it works
For each $x$, comparison with $\sum M_k$ shows that $\sum f_k(x)$ converges absolutely; call the sum $f(x)$. The tail is bounded the same way for every $x$: $$\left|f(x) - \sum_{k=1}^{n} f_k(x)\right| = \left|\sum_{k > n} f_k(x)\right| \le \sum_{k > n} M_k.$$ The right side does not depend on $x$ and goes to $0$ (it is the tail of a convergent series). So the partial sums converge uniformly. $\blacksquare$ Combined with the previous topic, a series of continuous functions that passes the M-test has a continuous sum, and can be integrated term by term.
The test is strong enough for a famous example. Weierstrass showed that $\sum_{k=0}^{\infty} \frac{1}{2^{k}}\cos(b^{k}\pi x)$, for a large odd integer $b$, is differentiable at no point at all. Yet the M-test with $M_k = \frac{1}{2^{k}}$ shows at once that the sum is continuous everywhere. Continuity survives uniform limits; differentiability does not.
Worked examples
✏️ Worked example: A trigonometric series
For $\sum_{k=1}^{\infty} \frac{\sin(kx)}{k^{2}}$ on $\mathbb{R}$, find $M_k$ and conclude.
- $\left|\frac{\sin(kx)}{k^{2}}\right| \le \frac{1}{k^{2}}$, with equality for some $x$.
- $\sum \frac{1}{k^{2}}$ converges.
Answer: $M_k = \frac{1}{k^{2}}$; the series converges uniformly on $\mathbb{R}$, so its sum is continuous.
✏️ Worked example: A power series on a smaller interval
For $\sum_{k=1}^{\infty} x^{k}$ on $\left[-\frac{2}{3}, \frac{2}{3}\right]$, find $\sum M_k$ and the smallest $N$ with tail bound $\sum_{k > N} M_k < 0.01$.
- $M_k = \left(\frac{2}{3}\right)^{k}$, so $\sum_{k \ge 1} M_k = \frac{2/3}{1/3} = 2$.
- The tail after $N$ terms is $\frac{(2/3)^{N+1}}{1/3} = 3\left(\frac{2}{3}\right)^{N+1}$, which first drops below $0.01$ at $N = 14$.
Answer: $\sum M_k = 2$; $N = 14$ works for every $x$ in the interval.
✏️ Worked example: Where the test fails
Can the M-test show that $\sum x^{k}$ converges uniformly on $(-1, 1)$?
- The best constant is $\sup_{(-1, 1)} |x^{k}| = 1$, and $\sum 1$ diverges.
Answer: No. (In fact the convergence is not uniform there: the partial sums blow up near $x = 1$.)
⚠️ Common mistake
An $M_k$ that depends on $x$, or one whose series diverges. "$|x^{k}| \le |x|^{k}$ and $\sum |x|^{k}$ converges" proves pointwise convergence only; the M-test needs numbers that work for all $x$ at once. And a bound whose series diverges proves nothing: the test simply does not apply, which is not the same as the series failing to converge.