The Weierstrass M-Test

The M-test is the standard engine behind proving that Convergence of Fourier Series define genuine continuous functions: coefficients like $a_n/n^2$ or $a_n/n^{1+\delta}$ (decaying fast enough) give an immediate convergent majorant $\sum |a_n|/n^2$, which yields uniform convergence of the Fourier series and hence continuity of its sum. In complex analysis, it underlies the theory of power series and analytic functions, guaranteeing that a power series converges uniformly on compact subsets of its disk of convergence, which is the key step in proving term-by-term differentiability of analytic functions. In numerical analysis and approximation theory, M-test–style dominated estimates justify swapping limits and integrals when constructing series solutions to differential equations, such as Fourier or Legendre series expansions used in solving the heat and wave equations. In probability theory, similar dominated-convergence-flavored arguments (though technically distinct, sharing the same spirit) justify interchanging sums and expectations for series of random variables. In cryptography and coding theory, generating functions and formal power series manipulations often rely on M-test-type uniform bounds to justify analytic continuations and asymptotic expansions used in complexity estimates.

:::quiz[Which of the following best explains why the counterexample series $f_n(x) = x^n(1-x)$ on $[0,1]$ cannot satisfy the M-test's hypotheses?]

• Because $f_n$ is not continuous at $x=1$
• Because the true supremum $\sup_x |f_n(x)| \sim 1/(en)$ forces any valid $M_n$ to form a divergent series like the harmonic series ✓
• Because $f_n(x)$ is negative for some $x$
• Because the pointwise limit function does not exist

Ask the tutor: How does uniform convergence, once established by the M-test, let us justify integrating a series term by term?

Think of each $f_n$ as a wave whose height at any point of $E$ never exceeds $M_n$, a fixed ceiling that does not depend on where you are standing in $E$. If the ceilings themselves form a convergent series — meaning that the total "budget" $M_{n_0} + M_{n_0+1} + \cdots$ shrinks to zero as $n_0 \to \infty$ — then no matter which point $x \in E$ you evaluate at, the tail of the function series is trapped under this shrinking budget. Since the budget shrinks uniformly, at the same rate for every $x$, the tails of the function series shrink uniformly too. This is exactly the content of the Cauchy Criterion for Uniform Convergence: uniform convergence holds precisely when the uniform norm of the tail, $\sup_x |\sum_{n=N}^{M} f_n(x)|$, can be forced below any $\varepsilon$ by choosing $N$ large, regardless of $M \ge N$ and regardless of $x$. The M-test hands you exactly this tail estimate for free, borrowed entirely from the numerical series $\sum M_n$.

:::svg[Each $f_n$ trapped in a shrinking band $[-M_n, M_n]$, forcing uniform smallness of the tail]

Example 1: a trigonometric series with non-symmetric coefficients. Consider, for $x \in \mathbb{R}$,

This is not a familiar named series, and there is no closed form for $f(x)$. We show it converges uniformly on all of $\mathbb{R}$. Let $f_n(x) = \dfrac{(-1)^n \sin(n^3 x)}{n^2+3n}$. Since $|\sin(n^3 x)| \le 1$ for every real $x$ and every $n$, and $|(-1)^n| = 1$,

Note $M_n$ does not depend on $x$ — this is the uniform bound hypothesis (i). Now check hypothesis (ii): compare $M_n$ to $1/n^2$. For $n \ge 1$,

and $\sum_{n=1}^\infty 1/n^2$ converges (a $p$-series with $p=2>1$), so by the comparison test $\sum_{n=1}^\infty M_n$ converges. Both hypotheses of the M-test hold on $E = \mathbb{R}$. Therefore $\sum_{n=1}^\infty f_n$ converges uniformly on $\mathbb{R}$, and $f$ is a well-defined function on $\mathbb{R}$. As a consequence — foreshadowing Uniform Convergence and Integration — since each $f_n$ is continuous and the convergence is uniform, $f$ is automatically continuous on $\mathbb{R}$, even though we have no closed-form expression for it.

Example 2: a series that is uniform on a restricted domain but not on all of $\mathbb{R}$. Consider, for $x \in \mathbb{R}$,

This is (up to sign and shift) related to the Taylor series of $-\ln(1 - x/2)$, but let us treat it directly as a series of functions $g_n(x) = \dfrac{x^n}{n \, 2^n}$ and ask on which sets the M-test applies. Fix a number $r$ with $0 < r < 2$, and take $E = [-r, r]$. For $x \in E$,

using $n \ge 1$ to drop the factor $1/n \le 1$ in the last inequality. Since $0 < r/2 < 1$, the series $\sum_{n=1}^\infty (r/2)^n$ is a convergent geometric series. Hence hypotheses (i) and (ii) hold on $E = [-r,r]$, and by the M-test $\sum g_n$ converges uniformly on $[-r, r]$ for every fixed $r < 2$. Now take $E = (-2, 2)$ itself, without restricting to a closed sub-interval. There is no single constant $M_n$ with $|g_n(x)| \le M_n$ for all $x \in (-2,2)$ and $\sum M_n < \infty$: as $x \to 2^-$, $\sup_{x \in (-2,2)} |g_n(x)| = 2^n/(n 2^n) = 1/n$, and $\sum 1/n$ diverges. So the M-test with $E = (-2,2)$ fails to apply — this does not prove non-uniform convergence, but it correctly signals that the natural majorant is unavailable, and indeed one can show directly (via the Cauchy criterion, evaluating near $x=2$) that the convergence fails to be uniform on the full open interval $(-2,2)$. This illustrates the typical use of the M-test for power series: uniform convergence is obtained on compact subintervals of the interval of convergence, not necessarily on the whole open interval.