Turning Majorant Bounds into a Proof
The Weierstrass M-Test turns a numerical comparison into a uniform-convergence result. The key is not merely that each function term is small: the bounds must be summable, and they must work at every point of the domain. We now prove that these conditions control both the pointwise series and the error in its partial sums.
Let \(E\) be a nonempty set, and let \(f_n:E\to\mathbb{R}\). Write \(S_N(x)=\sum_{n=1}^{N}f_n(x)\) for the partial sums. Suppose that \(M_n\geq0\), that \(|f_n(x)|\leq M_n\) for every \(n\geq1\) and \(x\in E\), and that \(\sum_{n=1}^{\infty}M_n\) converges. For \(N\geq0\), denote the numerical tail by $$ R_N=\sum_{n=N+1}^{\infty}M_n. $$ Convergence of the numerical series implies \(R_N\to0\). We will show that the same tail bounds the error of the function series at every point.
Proof. Fix \(x\in E\). For each \(n\), the assumed bound gives $$ 0\leq |f_n(x)|\leq M_n. $$ By comparison with the convergent nonnegative series \(\sum M_n\), the numerical series \(\sum |f_n(x)|\) converges. Thus \(\sum f_n(x)\) converges absolutely. This holds for every \(x\in E\), so the sum function $$ S(x)=\sum_{n=1}^{\infty}f_n(x) $$ is well defined on \(E\).
Now fix \(N\geq0\) and \(m>N\). The triangle inequality and the assumed bounds imply $$ |S_m(x)-S_N(x)| =\left|\sum_{n=N+1}^{m}f_n(x)\right| \leq\sum_{n=N+1}^{m}|f_n(x)| \leq\sum_{n=N+1}^{m}M_n \leq R_N. $$ For each fixed \(x\), we have \(S_m(x)\to S(x)\) as \(m\to\infty\). Taking this limit in the inequality gives $$ |S(x)-S_N(x)|\leq R_N \qquad\text{for every }x\in E. $$ Given \(\varepsilon>0\), choose \(N\) so that \(R_N<\varepsilon\), which is possible because \(R_N\to0\). Then for every \(x\in E\), $$ |S(x)-S_N(x)|\leq R_N<\varepsilon. $$ The same \(N\) works for all \(x\), so \(S_N\to S\) uniformly on \(E\). \(\square\)
The Uniform Remainder Estimate
The proof yields more than uniform convergence: it gives an explicit error bound for every truncation. This is often the most useful part of the test in applications. Once a majorant has been found, its numerical tail provides a uniform bound on the error, without requiring a formula for the function sum.
Proof. The proof of the M-Test established that \(|S(x)-S_N(x)|\leq R_N\) for every \(x\in E\), where \(R_N=\sum_{n=N+1}^{\infty}M_n\) is independent of \(x\). Taking the supremum over \(E\) preserves the inequality and gives the result. \(\square\)
This estimate also shows precisely how the numerical convergence controls uniform convergence. If the tail of \(\sum M_n\) is less than a chosen tolerance, then the error in the function series is below that tolerance everywhere on the domain. A useful special case occurs when the majorants decrease geometrically.
Proof. The geometric series \(\sum_{n=1}^{\infty}Cr^n\) converges when \(0\leq r<1\), so the Weierstrass M-Test applies with \(M_n=Cr^n\). The uniform remainder bound gives $$ \sup_{x\in E}|S(x)-S_N(x)| \leq\sum_{n=N+1}^{\infty}Cr^n. $$ If \(r=0\), every bound \(Cr^n\) is zero, so the remainder is zero. If \(0<r<1\), the geometric-series formula yields $$ \sum_{n=N+1}^{\infty}Cr^n =Cr^{N+1}\sum_{k=0}^{\infty}r^k =\frac{Cr^{N+1}}{1-r}. $$ Thus the stated estimate holds in both cases. \(\square\)
Worked Applications of the Proof
Worked Example: A Geometric Series of Functions on a Closed Interval
Consider $$ \sum_{n=1}^{\infty}\left(\frac{x}{4}\right)^n \qquad\text{on }[-2,2]. $$ For every \(x\in[-2,2]\), \(|x|\leq2\), so $$ \left|\left(\frac{x}{4}\right)^n\right| =\left(\frac{|x|}{4}\right)^n \leq\left(\frac12\right)^n. $$ The majorant series converges, since $$ \sum_{n=1}^{\infty}\left(\frac12\right)^n=1. $$ The M-Test therefore gives uniform and absolute convergence on \([-2,2]\). The geometric remainder estimate gives, for every \(N\geq0\), $$ \sup_{x\in[-2,2]} \left|\sum_{n=N+1}^{\infty}\left(\frac{x}{4}\right)^n\right| \leq\sum_{n=N+1}^{\infty}\left(\frac12\right)^n =\frac{(1/2)^{N+1}}{1-1/2} =\frac{1}{2^N}. $$ For instance, after \(N=4\) terms, the error is at most \(1/16\) throughout the whole interval.
Worked Example: Oscillation with a Reciprocal-Square Bound
Define \(f_n(x)=\cos(n^2x)/n^2\) for \(x\in\mathbb{R}\). Since \(|\cos t|\leq1\) for every real \(t\), $$ |f_n(x)|\leq\frac{1}{n^2} \qquad\text{for every }x\in\mathbb{R}. $$ The numerical series \(\sum 1/n^2\) converges, so the M-Test gives uniform and absolute convergence on all of \(\mathbb{R}\). An explicit tail estimate follows from the integral comparison: for \(N\geq1\), the function \(t\mapsto t^{-2}\) is decreasing and positive, and hence $$ \sum_{n=N+1}^{\infty}\frac{1}{n^2} \leq\int_N^\infty\frac{1}{t^2}\,dt =\left[-\frac1t\right]_{N}^{\infty} =\frac1N. $$ Consequently, $$ \sup_{x\in\mathbb{R}} \left|\sum_{n=N+1}^{\infty}\frac{\cos(n^2x)}{n^2}\right| \leq\frac1N. $$ Although the oscillation becomes increasingly rapid as \(n\) grows, it cannot make a term larger than its reciprocal-square bound.
Worked Example: A Rational Function Series Near Zero
On \(E=[0,1/3]\), consider $$ \sum_{n=1}^{\infty}\frac{x^n}{1+x^n}. $$ The denominator is positive, and \(0\leq x\leq1/3\), so \(0\leq x^n\leq3^{-n}\). Therefore $$ \left|\frac{x^n}{1+x^n}\right| =\frac{x^n}{1+x^n} \leq x^n \leq\frac{1}{3^n}. $$ The majorants sum to a convergent geometric series. The M-Test proves uniform and absolute convergence on \([0,1/3]\). More precisely, the error after \(N\) terms satisfies $$ \sup_{x\in[0,1/3]} \left|\sum_{n=N+1}^{\infty}\frac{x^n}{1+x^n}\right| \leq\sum_{n=N+1}^{\infty}\frac1{3^n} =\frac{(1/3)^{N+1}}{1-1/3} =\frac{1}{2\cdot3^N}. $$ The domain restriction is important: it supplies a geometric bound independent of \(x\).
What the Proof Does—and Does Not—Require
The proof has two logically distinct parts. First, fixing \(x\) turns the function terms into a numerical series, and comparison proves absolute convergence at that point. Second, the tail estimate has no \(x\) in it. That independence lets one choose a single \(N\) for the entire domain, which is exactly what uniform convergence requires.
A common mistake is to replace summability of the majorants by the weaker condition \(M_n\to0\). The latter only says that the individual terms are forced to become small; it does not make the tails \(\sum_{n>N}M_n\) small. For example, the numerical terms \(M_n=1/n\) tend to zero, but their series diverges. Thus a bound by \(1/n\) alone cannot invoke the M-Test.
It is also essential that the majorants be independent of \(x\). Bounds that vary with the input may still help prove convergence at each fixed point, but they do not automatically provide one error tolerance that works everywhere. Finally, the M-Test is sufficient rather than necessary: failure to find a summable majorant does not prove that uniform convergence fails. Its strength is that, once its hypotheses are verified, the proof supplies absolute convergence, uniform convergence, and a concrete error estimate together.
Check Your Understanding
Use the proof and estimates in this tutorial to answer the following questions.
- At what point in the proof is the assumption that \(\sum M_n\) converges used to establish absolute convergence at a fixed \(x\)?
- Why does the estimate for \(|S(x)-S_N(x)|\) prove uniform convergence rather than only pointwise convergence?
- For majorants \(M_n=Cr^n\) with \(0<r<1\), what is the uniform error bound after \(N\) terms?
- What uniform remainder bound follows from \(|f_n(x)|\leq1/n^2\) for every \(x\) and \(n\), when \(N\geq1\)?
- Why does knowing only that \(M_n\to0\) not permit an application of the M-Test?