From a Convergence Test to a Working Tool
The Weierstrass M-Test does more than certify that a series of functions converges uniformly. Its majorant gives a numerical error estimate that applies at every input. This makes the test useful when approximating a function, integrating a series, or establishing differentiability. In each application, the key question is what quantity needs to be bounded uniformly: the terms themselves, their integrals, or their derivatives.
We will use the Weierstrass M-Test and its Uniform Remainder Bound from the previous tutorials. We will also invoke two results established earlier in this course: uniform limits of Riemann integrable functions are Riemann integrable, with convergence of their integrals, and the Local Uniform Derivative Criterion. These results explain why uniform convergence matters; the M-Test supplies concrete hypotheses and quantitative estimates for applying them.
Uniform Approximation on an Unbounded Domain
A useful feature of the M-Test is that the domain need not be bounded. For example, suppose a function \(q\) satisfies \(|q(x)|\leq r<1\) for every \(x\) in its domain. Then the terms \(q(x)^n\) are bounded by the summable geometric sequence \(r^n\). The partial sums approximate the series uniformly, even if the domain is all of \(\mathbb{R}\).
Worked Example: A Rational Function as a Uniformly Convergent Series
For \(x\in\mathbb{R}\), set \(q(x)=x/(1+x^2)\), and consider $$ \sum_{n=1}^{\infty}q(x)^n. $$ The inequality \((|x|-1)^2\geq0\) gives \(x^2+1\geq2|x|\). Therefore $$ |q(x)|=\frac{|x|}{1+x^2}\leq\frac12 \qquad\text{for every }x\in\mathbb{R}. $$ Thus \(|q(x)^n|\leq2^{-n}\), and \(\sum_{n=1}^{\infty}2^{-n}\) converges. The M-Test gives uniform and absolute convergence on \(\mathbb{R}\). For each finite \(N\), the geometric-sum identity gives $$ \sum_{n=1}^{N}q(x)^n=\frac{q(x)(1-q(x)^N)}{1-q(x)}. $$ Since \(|q(x)|\leq1/2\), \(1-q(x)\neq0\), and the remainder satisfies $$ \left|\sum_{n=N+1}^{\infty}q(x)^n\right| =\left|\frac{q(x)^{N+1}}{1-q(x)}\right| \leq\frac{(1/2)^{N+1}}{1-1/2} =2^{-N}. $$ The pointwise geometric-series formula identifies the sum as $$ \sum_{n=1}^{\infty}q(x)^n =\frac{q(x)}{1-q(x)} =\frac{x}{x^2-x+1}. $$ The denominator is positive because \(x^2-x+1=(x-1/2)^2+3/4\). Hence the rational function is the uniform limit of the displayed partial sums, with error at most \(2^{-N}\) after \(N\) terms. Since each term is continuous, the continuity result for M-Test series also shows that the sum is continuous.
This example illustrates a practical approach to uniform approximation: find a simple bound on an auxiliary quantity, use it to produce a summable majorant, and then retain the majorant's tail as an explicit error estimate. The bound need not be sharp to be useful; it must be independent of the input and summable over the index.
Term-by-Term Integration with an Error Bound
On a compact interval, a summable uniform bound lets us integrate a function series term by term. The reason is that the M-Test gives uniform convergence of the partial sums, after which the theorem on convergence of Riemann integrals under uniform convergence applies. The same bound also controls how far the integral of a partial sum is from the integral of the full sum.
Proof. By the Weierstrass M-Test, \(S_N\) converges uniformly to \(S\) on \([a,b]\). Each \(S_N\) is Riemann integrable, since it is a finite sum of Riemann integrable functions. The earlier theorem on uniform limits of Riemann integrable functions therefore implies that \(S\) is Riemann integrable and that $$ \lim_{N\to\infty}\int_a^b S_N(x)\,dx=\int_a^b S(x)\,dx. $$ By linearity of the integral, \(\int_a^b S_N(x)\,dx=\sum_{n=1}^{N}\int_a^b f_n(x)\,dx\). Taking the limit proves the claimed equality of the integral and the series of integrals.
For the error estimate, the Uniform Remainder Bound gives \(|S(x)-S_N(x)|\leq\sum_{n=N+1}^{\infty}M_n\) at every \(x\in[a,b]\). The integral inequality for Riemann integrable functions now yields $$ \left|\int_a^b(S(x)-S_N(x))\,dx\right| \leq\int_a^b|S(x)-S_N(x)|\,dx \leq(b-a)\sum_{n=N+1}^{\infty}M_n. $$ Linearity identifies the left-hand side with the asserted integral error. If \(a=b\), both integrals are zero and the same inequality holds. \(\square\)
Worked Example: Integrating a Series on a Short Interval
On \([0,1/2]\), consider $$ \sum_{n=1}^{\infty}\frac{x^n}{n+1}. $$ Each term is continuous, and $$ \left|\frac{x^n}{n+1}\right|\leq\frac{2^{-n}}{n+1}\leq2^{-n} \qquad (0\leq x\leq1/2). $$ Because \(\sum 2^{-n}\) converges, the series converges uniformly on this interval. The integral tail bound applies with \(M_n=2^{-n}\). Direct integration of each term gives $$ \int_0^{1/2}\frac{x^n}{n+1}\,dx =\frac{(1/2)^{n+1}}{(n+1)^2}. $$ Consequently, $$ \int_0^{1/2}\left(\sum_{n=1}^{\infty}\frac{x^n}{n+1}\right)\,dx =\sum_{n=1}^{\infty}\frac{(1/2)^{n+1}}{(n+1)^2}. $$ After integrating the first \(N\) terms, the error is bounded by $$ \frac12\sum_{n=N+1}^{\infty}2^{-n} =\frac12\cdot2^{-N} =2^{-(N+1)}. $$ The bound does not require finding a closed form for the function sum or for the resulting numerical series.
Derivative Bounds and Differentiability
For differentiation, bounding the terms \(f_n\) is not the central requirement. Instead, one seeks a summable bound on the derivatives \(f_n'\). The Local Uniform Derivative Criterion then turns uniform convergence of the derivative partial sums into differentiability of the original sum, provided the function partial sums converge at at least one point. The pointwise condition matters: control of the derivatives alone does not determine the values of the functions, since constants disappear upon differentiation.
Proof. The bounds on \(f_n'\) and the M-Test show that the derivative partial sums \(\sum_{n=1}^{N}f_n'\) converge uniformly on \(J\). The partial sums \(S_N=\sum_{n=1}^{N}f_n\) are continuously differentiable, and \(S_N(x_0)\) converges by assumption. The Local Uniform Derivative Criterion therefore gives differentiability of \(S\) and the stated derivative formula, with uniform convergence of the derivative series.
Fix \(N\geq0\), and define \(G=S-S_N\). The derivative formula gives $$ G'(t)=\sum_{n=N+1}^{\infty}f_n'(t), \qquad |G'(t)|\leq\sum_{n=N+1}^{\infty}M_n \qquad(t\in J). $$ For any \(x,y\in J\), the closed segment joining \(x\) and \(y\) lies in \(J\). If \(x\neq y\), the Mean Value Theorem applied to \(G\) on that segment gives a point \(c\) between \(x\) and \(y\) such that \(|G(x)-G(y)|=|x-y||G'(c)|\). The derivative bound proves the desired inequality. If \(x=y\), its left-hand side is zero, so the inequality holds as well. \(\square\)
The proposition provides more information than differentiability: the derivative tail controls how quickly the function tail can vary across the interval. It does not by itself bound the value of that tail at a particular point. To bound the values, one can combine it with a bound at a chosen base point, or separately apply the M-Test to the original terms.
Worked Example: Differentiating an Oscillatory Series
For \(x\in\mathbb{R}\), define $$ S(x)=\sum_{n=1}^{\infty}\frac{\sin(nx)}{n^3}. $$ The terms are bounded by \(1/n^3\), so the M-Test gives uniform convergence on \(\mathbb{R}\). To study derivatives, set \(f_n(x)=\sin(nx)/n^3\). Then $$ f_n'(x)=\frac{\cos(nx)}{n^2}, \qquad |f_n'(x)|\leq\frac1{n^2}. $$ The majorants \(1/n^2\) are summable. At \(x_0=0\), every \(f_n(0)=0\), so the series of values converges there. The proposition gives $$ S'(x)=\sum_{n=1}^{\infty}\frac{\cos(nx)}{n^2}, $$ and this derivative series converges uniformly on \(\mathbb{R}\). In particular, for \(N\geq1\), integral comparison gives $$ \sup_{x\in\mathbb{R}}\left|S'(x)-\sum_{n=1}^{N}\frac{\cos(nx)}{n^2}\right| \leq\sum_{n=N+1}^{\infty}\frac1{n^2} \leq\int_N^\infty\frac{dt}{t^2} =\frac1N. $$ The original series also has a uniform value-error bound: $$ \sup_{x\in\mathbb{R}}\left|S(x)-\sum_{n=1}^{N}\frac{\sin(nx)}{n^3}\right| \leq\sum_{n=N+1}^{\infty}\frac1{n^3} \leq\int_N^\infty\frac{dt}{t^3} =\frac1{2N^2}. $$ Thus the same truncation gives explicit, uniform estimates for both the function and its derivative.
Choosing the Right Majorant
These applications use the M-Test in related but distinct ways. For uniform approximation, majorize the terms \(f_n(x)\). For integration, the same bound controls the error after integrating because the interval has finite length. For differentiation, majorize \(f_n'(x)\) and also check convergence at one point. A bound on the derivatives cannot replace that final check: for instance, constant functions can have zero derivatives while their series of values diverges.
A summable majorant is a sufficient condition, not a necessary one. If no convenient majorant is available, that does not establish failure of uniform convergence or failure of term-by-term operations. But when a majorant is available, it gives a useful package: convergence, uniform control, and a numerical estimate whose tail can be chosen to meet a desired tolerance.
Check Your Understanding
Use the applications and estimates in this tutorial to answer the following questions.
- Why does the bound \(|x/(1+x^2)|\leq1/2\) give uniform convergence of its geometric function series on all of \(\mathbb{R}\)?
- For an M-Test series on \([a,b]\), what factor converts the uniform remainder bound into the integral error bound?
- In the derivative-tail proposition, why is convergence of \(\sum f_n(x_0)\) required at one point?
- For the series \(\sum_{n=1}^{\infty}\sin(nx)/n^3\), what uniform error bounds hold for the function and derivative partial sums after \(N\geq1\) terms?
- Does failure to find a summable majorant prove that a function series is not uniformly convergent? Explain.