Prove the Weierstrass M-Test

From Numerical Series to Function Series

In the previous tutorial, we proved that absolute convergence controls the tails of a numerical series and guarantees convergence. The Weierstrass M-Test transfers this idea to a series of functions: if the absolute value of each function is bounded by a term in a convergent numerical series, then the function series converges uniformly. The essential feature is that the same numerical bound works at every point of the domain.

Uniform convergence is stronger than pointwise convergence. Pointwise convergence allows the index after which an error is small to depend on the point $x$. Uniform convergence requires one index to work for every $x$ at once. A summable sequence of bounds supplies exactly this common control.

Definition: Let $E$ be a set, and let $f_n:E\to\mathbb R$ for each positive integer $n$. Define the $N$th partial-sum function by $$ S_N(x)=\sum_{n=1}^{N}f_n(x). $$ The function series $\sum_{n=1}^{\infty}f_n(x)$ converges uniformly on $E$ to a function $S:E\to\mathbb R$ if, for every $\varepsilon>0$, there is a positive integer $N$ such that $$ |S_N(x)-S(x)|<\varepsilon $$ for every $x\in E$ and every integer $N$ at least that large.

For each fixed $x$, the expression $\sum_{n=1}^{\infty}f_n(x)$ is an ordinary numerical series. The M-Test first ensures that this numerical series converges absolutely for every $x$, so it defines a function $S$. It then proves that the convergence to $S$ is uniform, not merely pointwise.

The Weierstrass M-Test

Theorem (Weierstrass M-Test): Let $E$ be a set, and let $f_n:E\to\mathbb R$ for each positive integer $n$. Suppose there are nonnegative real numbers $M_n$ such that $$ |f_n(x)|\leq M_n $$ for every $x\in E$ and every $n$, and suppose the numerical series $\sum_{n=1}^{\infty}M_n$ converges. Then $\sum_{n=1}^{\infty}f_n(x)$ converges uniformly on $E$. In fact, it converges absolutely at every $x\in E$.

Proof. If $E$ is empty, the assertion of uniform convergence is vacuous. Suppose $E$ is nonempty. Fix $x\in E$. The assumed bound gives $|f_n(x)|\leq M_n$ for every $n$. Since $\sum M_n$ converges and has nonnegative terms, the Comparison Test shows that $\sum |f_n(x)|$ converges. Thus $\sum f_n(x)$ converges absolutely, by the Absolute Convergence Implies Convergence theorem. We can therefore define

S(x)=\sum_{n=1}^{\infty}f_n(x)

for every $x\in E$.

It remains to prove uniform convergence. Let $\varepsilon>0$. Since $\sum M_n$ converges, the Cauchy Criterion for Series gives a positive integer $N_0$ such that

\sum_{k=n+1}^{m}M_k<\varepsilon

whenever $m>n\geq N_0$. For any $x\in E$ and $m>n\geq N_0$, the finite triangle inequality and the majorant bounds yield

\left|\sum_{k=n+1}^{m}f_k(x)\right| \leq \sum_{k=n+1}^{m}|f_k(x)| \leq \sum_{k=n+1}^{m}M_k <\varepsilon.

The final bound does not depend on $x$. Consequently, for any $p,q\geq N_0$, the difference $|S_p(x)-S_q(x)|$ is less than $\varepsilon$, for every $x\in E$: if $p>q$, apply the displayed estimate with $n=q$ and $m=p$; if $q>p$, reverse the difference; and if $p=q$, the difference is zero. The partial-sum functions are therefore uniformly Cauchy. By the Uniform Cauchy Criterion, they converge uniformly on $E$ to a real-valued function. Their pointwise limits are the sums $S(x)$ already defined, so the uniform limit is $S$. This proves the theorem. $\square$

The proof has two distinct tasks. Comparison gives convergence at each fixed point, while the bound on every finite tail proves the uniform Cauchy condition. The latter depends on the majorants being independent of $x$; without that common bound, pointwise convergence alone does not provide uniform convergence.

A Uniform Bound on the Remainder

The same estimates give more than uniform convergence. They quantify how far the sum can be from a chosen partial sum. This is the function-series analogue of the Absolute-Convergence Tail Estimate from the previous tutorial.

Theorem (Uniform Remainder Estimate): Under the hypotheses of the Weierstrass M-Test, let $$ S(x)=\sum_{n=1}^{\infty}f_n(x). $$ Then, for every positive integer $N$ and every $x\in E$, $$ |S(x)-S_N(x)|\leq \sum_{k=N+1}^{\infty}M_k. $$

Proof. Fix $N$ and $x\in E$. For every $m>N$, the finite triangle inequality gives

\left|\sum_{k=N+1}^{m}f_k(x)\right| \leq \sum_{k=N+1}^{m}|f_k(x)| \leq \sum_{k=N+1}^{m}M_k.

As $m\to\infty$, the sum inside the absolute value on the left tends to $S(x)-S_N(x)$, because the first $N$ terms have been subtracted from the partial sums. By the Absolute-Value Limit Theorem, its absolute value tends to $|S(x)-S_N(x)|$. The right side tends to $\sum_{k=N+1}^{\infty}M_k$. Order is preserved under limits, so passing to the limit proves the estimate. $\square$

Because the right side is independent of $x$, the estimate controls the error over the entire domain. Given a desired error $\varepsilon>0$, it is enough to choose $N$ so that the numerical tail $\sum_{k=N+1}^{\infty}M_k$ is less than $\varepsilon$. The resulting error bound then holds simultaneously at every point.

Worked Applications

Worked Example: A Geometric Function Series on a Closed Interval

On $E=[-1,1]$, define $f_n(x)=x^n/5^n$. Since $|x|\leq1$,

|f_n(x)|=\frac{|x|^n}{5^n}\leq\frac{1}{5^n}.

The numerical majorant series is geometric and converges:

\sum_{n=1}^{\infty}\frac{1}{5^n} =\frac{\frac15}{1-\frac15} =\frac14.

The Weierstrass M-Test proves that $\sum_{n=1}^{\infty}x^n/5^n$ converges uniformly on $[-1,1]$. The uniform remainder estimate is

|S(x)-S_N(x)| \leq\sum_{k=N+1}^{\infty}\frac1{5^k} =\frac{5^{-(N+1)}}{1-\frac15} =\frac{1}{4\cdot5^N}.

For instance, $N=3$ gives an error at most $1/(4\cdot125)=1/500$ at every $x\in[-1,1]$. The estimate is valid at the endpoints as well as in the interior.

Worked Example: Trigonometric Terms on the Real Line

Let $E=\mathbb R$ and set $f_n(x)=\sin(nx)/[n(n+1)]$. The bound $|\sin(nx)|\leq1$ gives

|f_n(x)|\leq\frac{1}{n(n+1)} =\frac1n-\frac{1}{n+1}.

The majorant series converges by telescoping: its $m$th partial sum is

\sum_{n=1}^{m}\left(\frac1n-\frac{1}{n+1}\right) =1-\frac{1}{m+1},

which tends to $1$. Hence the M-Test proves uniform convergence of $\sum_{n=1}^{\infty}\sin(nx)/[n(n+1)]$ on all of $\mathbb R$. For every $N\geq1$, the error satisfies

|S(x)-S_N(x)| \leq\sum_{k=N+1}^{\infty}\left(\frac1k-\frac{1}{k+1}\right) =\frac{1}{N+1}.

Thus the tails are uniformly small even though the domain is unbounded. Compactness of the domain is not a hypothesis of the M-Test.

Worked Example: A Power Series on the Unit Interval

On $E=[0,1]$, let $f_n(x)=x^n/n^2$. Since $0\leq x\leq1$, we have

|f_n(x)|=\frac{x^n}{n^2}\leq\frac1{n^2}.

The majorant series converges. Indeed, for $n\geq2$,

\frac1{n^2}\leq\frac1{n(n-1)} =\frac1{n-1}-\frac1n.

The sum of the right-hand side from $n=2$ to $m$ is $1-1/m$, so the partial sums of $\sum 1/n^2$ are bounded above by $2$. They are nondecreasing because their terms are positive; the Monotone Convergence Theorem therefore gives convergence. The M-Test now proves that $\sum_{n=1}^{\infty}x^n/n^2$ converges uniformly on $[0,1]$.

There is also a convenient explicit tail bound. For $N\geq1$, the same comparison gives

|S(x)-S_N(x)| \leq\sum_{n=N+1}^{\infty}\frac1{n^2} \leq\sum_{n=N+1}^{\infty}\left(\frac1{n-1}-\frac1n\right) =\frac1N.

In particular, choosing $N=100$ guarantees an error of at most $1/100$ everywhere on the interval.

What the Majorant Must Accomplish

A common mistake is to find a bound for each function without checking whether the bounds form a convergent series. For example, on $[0,1]$, the functions $f_n(x)=x^n$ satisfy $|f_n(x)|\leq1$. But the proposed constant majorants $M_n=1$ do not have a convergent sum. The M-Test therefore gives no conclusion. In fact, at $x=1$, the series becomes $\sum_{n=1}^{\infty}1$, which diverges. This example shows why the summability condition matters; a pointwise bound alone is not enough.

A successful application follows a short sequence of checks: identify the domain, obtain bounds that hold for every point of that domain, verify that the numerical series of bounds converges, and use its tails to control the function-series tails. The bounds need not be sharp. A simple summable majorant is often more useful than the smallest possible one.

Choose a common majorant.
Find nonnegative numbers $M_n$ such that $|f_n(x)|\leq M_n$ for every $x$ in the domain.

Verify numerical convergence.
Prove that $\sum M_n$ converges; this is the condition that makes its tails arbitrarily small.

Control all function tails.
Apply the finite triangle inequality and then the uniform bound on the majorant tail.

Conclude uniform convergence.
The partial sums are uniformly Cauchy, and the Uniform Cauchy Criterion supplies their uniform limit.

Check Your Understanding

Use the definition, theorem, and examples above to answer these questions.

Why does the comparison with $\sum M_n$ imply absolute convergence at each fixed $x$?
Which part of the M-Test proof makes its Cauchy estimate uniform in $x$?
For $f_n(x)=\cos(nx)/n^2$ on $\mathbb R$, what majorant would you try, and what fact would you need to verify about its series?
How does the Uniform Remainder Estimate help choose a partial sum to meet a prescribed error tolerance?
Why does the bound $|x^n|\leq1$ on $[0,1]$ not by itself establish uniform convergence of $\sum x^n$?

What You'll Learn