Technopedia

The technologist’s encyclopedia

Sequences and series

Sequences and series provide the tools for deciding when an infinite sum converges, how fast it converges, and what function it defines. They are the infinite-process layer of real analysis: power series turn local derivative data into global function representation, and uniform convergence of series guarantees that the limit function inherits the properties (continuity, differentiability, integrability) of its terms. This page covers the standard convergence tests, power series and their radius of convergence, and the Weierstrass M-test for uniform convergence of series.

Authoritative sources

Convergence tests

Definition. A series ∑n=1∞an\sum_{n=1}^{\infty} a_n converges if the sequence of partial sums sN=∑n=1Nans_N = \sum_{n=1}^{N} a_n converges to a finite limit.

Theorem (ratio test). If lim⁡n→∞∣an+1/an∣=L\lim_{n \to \infty} |a_{n+1}/a_n| = L, then the series converges absolutely if L<1L < 1 and diverges if L>1L > 1.

Theorem (root test). If lim⁡n→∞∣an∣n=L\lim_{n \to \infty} \sqrt[n]{|a_n|} = L, then the series converges absolutely if L<1L < 1 and diverges if L>1L > 1.

Theorem (comparison test). If 0≤an≤bn0 \leq a_n \leq b_n for all nn and ∑bn\sum b_n converges, then ∑an\sum a_n converges.

Theorem (integral test). If f:[1,∞)→[0,∞)f: [1, \infty) \to [0, \infty) is decreasing and an=f(n)a_n = f(n), then ∑an\sum a_n converges if and only if ∫1∞f(x) dx\int_1^{\infty} f(x)\, dx converges.

Details

Why tests are needed

The definition of convergence — “the partial sums approach a finite limit” — is rarely checkable directly for an infinite series. Tests give sufficient (or necessary-and-sufficient) conditions in terms of the growth rate of the terms, which is much easier to compute.

The ratio test in action

Apply the ratio test to ∑n=1∞xnn!\sum_{n=1}^{\infty} \frac{x^n}{n!}.

∣an+1an∣=xn+1/(n+1)!xn/n!=∣x∣n+1→0<1.\left|\frac{a_{n+1}}{a_n}\right| = \frac{x^{n+1}/(n+1)!}{x^n/n!} = \frac{|x|}{n+1} \to 0 < 1.\n\nSo the series converges absolutely for every x∈Rx \in \mathbb{R}. (This is the series for exe^x.)

The comparison test in action

Compare ∑1n2\sum \frac{1}{n^2} with the telescoping series. Note that for n≥2n \geq 2, 1n2<1n(n−1)=1n−1−1n\frac{1}{n^2} < \frac{1}{n(n-1)} = \frac{1}{n-1} - \frac{1}{n}. The right side telescopes: ∑n=2N(1n−1−1n)=1−1N→1\sum_{n=2}^{N} \left(\frac{1}{n-1} - \frac{1}{n}\right) = 1 - \frac{1}{N} \to 1. Hence ∑1n2\sum \frac{1}{n^2} converges.

The integral test in action

For ∑1nln⁡n\sum \frac{1}{n \ln n} (starting at n=2n = 2): set f(x)=1/(xln⁡x)f(x) = 1/(x \ln x). Then ∫2∞dxxln⁡x=[ln⁡ln⁡x]2∞=∞\int_2^{\infty} \frac{dx}{x \ln x} = [\ln \ln x]_2^{\infty} = \infty, so the series diverges.

Example problem

Problem. Determine whether ∑n=1∞n2n\sum_{n=1}^{\infty} \frac{n}{2^n} converges, and if so, find its sum.

Solution

Convergence. Ratio test: an+1an=n+12n+1⋅2nn=n+12n→12<1\frac{a_{n+1}}{a_n} = \frac{n+1}{2^{n+1}} \cdot \frac{2^n}{n} = \frac{n+1}{2n} \to \frac{1}{2} < 1. Converges.

Sum. Let S=∑n=1∞n2n=∑n=1∞nxnS = \sum_{n=1}^{\infty} \frac{n}{2^n} = \sum_{n=1}^{\infty} n x^n with x=1/2x = 1/2. We know ∑n=0∞xn=11−x\sum_{n=0}^{\infty} x^n = \frac{1}{1-x} for ∣x∣<1|x| < 1. Differentiating: ∑n=1∞nxn−1=1(1−x)2\sum_{n=1}^{\infty} n x^{n-1} = \frac{1}{(1-x)^2}, so ∑n=1∞nxn=x(1−x)2=1/2(1/2)2=2\sum_{n=1}^{\infty} n x^n = \frac{x}{(1-x)^2} = \frac{1/2}{(1/2)^2} = 2. ∎

Power series

Definition. A power series centered at aa is ∑n=0∞cn(x−a)n\sum_{n=0}^{\infty} c_n (x - a)^n. It has a radius of convergence R∈[0,∞]R \in [0, \infty] such that it converges absolutely for ∣x−a∣<R|x - a| < R and diverges for ∣x−a∣>R|x - a| > R.

Theorem (radius of convergence). If L=lim sup⁡n→∞∣cn∣nL = \limsup_{n \to \infty} \sqrt[n]{|c_n|}, then R=1/LR = 1/L (with the conventions 1/0=∞1/0 = \infty and 1/∞=01/\infty = 0).

Theorem (term-by-term differentiation). Inside its radius of convergence, a power series may be differentiated and integrated term by term, and the resulting series has the same radius of convergence.

Details

Why power series matter

A power series is the most general form of a function that is “infinitely differentiable and locally determined by its derivatives at one point.” The Taylor series of a function is its power series with cn=f(n)(a)/n!c_n = f^{(n)}(a)/n!. The deep question — when does the Taylor series actually converge to the function? — is answered by the remainder estimate in Taylor’s theorem (see Differentiation).

Worked example: geometric series

∑n=0∞xn=11−x\sum_{n=0}^{\infty} x^n = \frac{1}{1-x} for ∣x∣<1|x| < 1. Here cn=1c_n = 1, so L=1L = 1 and R=1R = 1. At the boundary x=1x = 1 the series diverges; at x=−1x = -1 it oscillates.

Worked example: ln⁡(1+x)\ln(1+x)

For ∣x∣<1|x| < 1: ln⁡(1+x)=∑n=1∞(−1)n+1xnn\ln(1+x) = \sum_{n=1}^{\infty} (-1)^{n+1} \frac{x^n}{n}. This is obtained by integrating the geometric series 11+t=∑tn\frac{1}{1+t} = \sum t^n term by term from 00 to xx. The radius is R=1R = 1 (the singularity at x=−1x = -1).

Term-by-term differentiation

Since 11−x=∑xn\frac{1}{1-x} = \sum x^n, differentiating gives 1(1−x)2=∑n=1∞nxn\frac{1}{(1-x)^2} = \sum_{n=1}^{\infty} n x^n for ∣x∣<1|x| < 1. This is the identity used in the ratio-test example above.

Example problem

Problem. Find the radius of convergence of ∑n=0∞(x−3)nn2+1\sum_{n=0}^{\infty} \frac{(x-3)^n}{n^2 + 1}.

Solution

cn=1n2+1c_n = \frac{1}{n^2+1}. By the root test, ∣cn∣n=(n2+1)−1/n→1\sqrt[n]{|c_n|} = (n^2+1)^{-1/n} \to 1 (since n2/n→1n^{2/n} \to 1). So L=1L = 1 and R=1R = 1. The series converges for ∣x−3∣<1|x - 3| < 1, i.e. x∈(2,4)x \in (2, 4). At the endpoints x=2x = 2 and x=4x = 4, the series becomes ∑(−1)nn2+1\sum \frac{(-1)^n}{n^2+1} and ∑1n2+1\sum \frac{1}{n^2+1}, both of which converge by comparison with ∑1/n2\sum 1/n^2. So the interval of convergence is [2,4][2, 4]. ∎

Uniform convergence of series

Definition. A series ∑fn\sum f_n of functions on XX converges uniformly if the sequence of partial sums sN=∑n=1Nfns_N = \sum_{n=1}^{N} f_n converges uniformly on XX.

Theorem (Weierstrass M-test). If ∣fn(x)∣≤Mn|f_n(x)| \leq M_n for all x∈Xx \in X and ∑Mn\sum M_n converges, then ∑fn\sum f_n converges uniformly (and absolutely) on XX.

Theorem (uniform limit of continuous functions). If fnf_n are continuous on XX and ∑fn\sum f_n converges uniformly, then the sum f=∑fnf = \sum f_n is continuous on XX.

Theorem (term-by-term integration). If fnf_n are Riemann integrable on [a,b][a,b] and ∑fn\sum f_n converges uniformly, then ∫ab∑fn=∑∫abfn\int_a^b \sum f_n = \sum \int_a^b f_n.

Details

Why uniform convergence of series matters

Pointwise convergence of a series of continuous functions can produce a discontinuous limit (as with fn(x)=xnf_n(x) = x^n on [0,1][0,1] — see Limits and continuity). Uniform convergence is the condition that prevents this: it guarantees the limit function is continuous, and it justifies interchanging the sum with integration and (under additional hypotheses) differentiation.

The M-test in action

Show that ∑n=1∞sin⁡(nx)n2\sum_{n=1}^{\infty} \frac{\sin(nx)}{n^2} converges uniformly on R\mathbb{R}.

Take Mn=1/n2M_n = 1/n^2 since ∣sin⁡(nx)∣≤1|\sin(nx)| \leq 1. The series ∑1/n2\sum 1/n^2 converges (p-series with p=2>1p = 2 > 1), so the M-test gives uniform convergence on all of R\mathbb{R}.

When the M-test fails but uniform convergence holds

The series ∑xnn2\sum \frac{x^n}{n^2} on [0,1][0, 1] converges uniformly (the sum is continuous on [0,1][0,1] and the convergence is monotone by Dini’s theorem), but the M-test with Mn=1/n2M_n = 1/n^2 works here too. A case where the M-test fails: ∑xnn\sum \frac{x^n}{n} on [0,1)[0, 1) — the M-test would need Mn=1/nM_n = 1/n and ∑1/n\sum 1/n diverges, yet the series converges uniformly on any [0,r][0, r] with r<1r < 1.

Term-by-term integration example

∫01∑n=1∞xnn2 dx=∑n=1∞1n2∫01xn dx=∑n=1∞1n2(n+1)=∑n=1∞(1n2−1n2(n+1))\int_0^1 \sum_{n=1}^{\infty} \frac{x^n}{n^2}\, dx = \sum_{n=1}^{\infty} \frac{1}{n^2} \int_0^1 x^n\, dx = \sum_{n=1}^{\infty} \frac{1}{n^2(n+1)} = \sum_{n=1}^{\infty} \left(\frac{1}{n^2} - \frac{1}{n^2(n+1)}\right).

The interchange is justified by uniform convergence (M-test with Mn=1/n2M_n = 1/n^2).

Example problem

Problem. Show that ∑n=1∞xnn\sum_{n=1}^{\infty} \frac{x^n}{n} converges uniformly on [0,r][0, r] for any r<1r < 1, but not on [0,1][0, 1].

Solution

On [0,r][0, r]: ∣xn/n∣≤rn/n≤rn|x^n/n| \leq r^n/n \leq r^n. Since ∑rn\sum r^n converges (geometric, r<1r < 1), the M-test gives uniform convergence.

Not on [0,1][0, 1]: The sum is −ln⁡(1−x)-\ln(1-x), which is unbounded as x→1−x \to 1^-. A uniform limit of continuous functions on [0,1][0,1] must be bounded (each partial sum is bounded, and uniform convergence preserves boundedness). Since −ln⁡(1−x)-\ln(1-x) is unbounded on [0,1][0,1], the convergence cannot be uniform. ∎

Quiz

Example problem

Problem. Consider the series ∑n=1∞(−1)n+1xnn\sum_{n=1}^{\infty} \frac{(-1)^{n+1} x^n}{n}.

(a) Find its radius of convergence.

(b) Identify the function it represents for ∣x∣<1|x| < 1.

(c) Show that it converges uniformly on [−r,r][-r, r] for any r<1r < 1.

(d) Does it converge uniformly on [−1,1][-1, 1]? Justify.

Solution

(a) cn=(−1)n+1/nc_n = (-1)^{n+1}/n, so ∣cn∣n=1/nn→1\sqrt[n]{|c_n|} = 1/\sqrt[n]{n} \to 1. Hence R=1R = 1.

(b) Differentiate term by term: ∑n=1∞(−1)n+1xn−1=∑n=0∞(−x)n=11+x\sum_{n=1}^{\infty} (-1)^{n+1} x^{n-1} = \sum_{n=0}^{\infty} (-x)^n = \frac{1}{1+x} for ∣x∣<1|x| < 1. Integrating from 00 to xx: ln⁡(1+x)=∑n=1∞(−1)n+1xnn\ln(1+x) = \sum_{n=1}^{\infty} \frac{(-1)^{n+1} x^n}{n}.

(c) On [−r,r][-r, r] with r<1r < 1: ∣(−1)n+1xn/n∣≤rn/n≤rn|(-1)^{n+1} x^n / n| \leq r^n / n \leq r^n. Since ∑rn\sum r^n converges, the M-test gives uniform convergence.

(d) At x=1x = 1, the series is the alternating harmonic series ∑(−1)n+1/n=ln⁡2\sum (-1)^{n+1}/n = \ln 2. At x=−1x = -1, it is −∑1/n=−∞-\sum 1/n = -\infty (diverges). So the series does not even converge at x=−1x = -1, and therefore cannot converge uniformly on [−1,1][-1, 1]. ∎