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Infinite Series of Functions

Definition

Consider a sequence of functions (fn)n=1∞(f_n)_{n=1}^\infty from XX to R.\mathbb{R}\text{.} If the sequence of partial sums SN(x)=∑n=1Nfn(x)\displaystyle S_N(x) = \sum_{n=1}^N f_n(x) converges to f ⁣:X→Rf\colon X \to \mathbb{R} pointwise or uniformly on XX as N→∞,N \to \infty\text{,} then we say the infinite series ∑n=1∞fn\displaystyle \sum_{n=1}^\infty f_n converges pointwise/uniformly to ff on X,X\text{,} denoted as f(x)=∑n=1∞fn(x).\displaystyle f(x) = \sum_{n=1}^\infty f_n(x)\text{.}


Interpretation

A series of functions is defined through its partial sums: each SNS_N is an ordinary finite sum of functions, and the series converges to ff when the sequence (SN)(S_N) does. Every question about a series of functions is therefore a question about the sequence of partial sums, and the results on pointwise and uniform convergence of sequences apply directly.

Partial sums of the series of sin(nx) over n squared, with 1, 2, 4 and 10 terms, on the interval from 0 to 2 pi; the later sums are almost indistinguishable.

The figure shows the partial sums with 11, 22, 44 and 1010 terms of ∑n≥1sin⁡(nx)/n2\displaystyle \sum_{n\ge1}\sin(nx)/n^2 on [0,2π][0,2\pi]. Each new term changes the curve less, and the later sums are almost indistinguishable. Since ∣sin⁡(nx)/n2∣≤1/n2|\sin(nx)/n^2|\le1/n^2 for every xx, the sum with 1010 terms differs from the full sum by at most ∑n>101/n2≈0.095\displaystyle \sum_{n>10}1/n^2\approx0.095 at every point at once; this is the bound used in the Weierstrass MM-test.

Pointwise convergence at xx is convergence of the numerical infinite series ∑nfn(x)\displaystyle \sum_n f_n(x), so the tests for numerical series apply for each fixed xx. Uniform convergence asks more: the largest error sup⁡x∈X∣f(x)−SN(x)∣\sup_{x\in X}|f(x)-S_N(x)| must tend to 00. A finite sum can always be differentiated and integrated term by term; for an infinite sum, the Key Results need uniform convergence. Without it, a sum of continuous functions can be discontinuous: ∑n≥0x2/(1+x2)n\displaystyle \sum_{n\ge0}x^2/(1+x^2)^n equals 1+x21+x^2 for x≠0x\ne0 and 00 at x=0x=0.

Power series are the most important special case, and normal convergence is the usual way to show that a series converges uniformly.

A common slip is to differentiate term by term because the series itself converges uniformly. The condition is on the derivative series: ∑nsin⁡(nx)/n2\displaystyle \sum_n\sin(nx)/n^2 converges uniformly, but its derivative series ∑ncos⁡(nx)/n\displaystyle \sum_n\cos(nx)/n is the harmonic series at x=0x=0 and diverges.


Key Results

  • If fnf_n's are (continuously) differentiable, ∑n=1∞fn→f,\displaystyle \sum_{n=1}^\infty f_n \rightarrow f\text{,} and ∑n=1∞fn′⇉g\displaystyle \sum_{n=1}^\infty f'_n\rightrightarrows g on (a,b),(a,b)\text{,} then ff is (continuously) differentiable on (a,b)(a, b) with f′=g,f' = g\text{,} i.e., for every x0∈(a,b)x_0\in(a,b),
∑n=1∞fn′(x0)=(∑n=1∞fn)′(x0).\sum_{n=1}^\infty f'_n(x_0)=\left(\sum_{n=1}^\infty f_n\right)'(x_0).
  • On a finite closed interval [a,b][a,b] with a<ba<b, if every fnf_n is Riemann integrable and ∑n=1∞fn\displaystyle \sum_{n=1}^\infty f_n converges uniformly, then its sum is Riemann integrable and
∫ab(∑n=1∞fn(x)) dx=∑n=1∞∫abfn(x) dx.\int_a^b \left(\sum_{n=1}^\infty f_n(x)\right) \, \mathrm{d}x = \sum_{n=1}^\infty \int_a^b f_n(x) \, \mathrm{d}x.

The finite interval matters: the integral of a uniform error is at most (b−a)(b-a) times that error. Uniform convergence alone does not provide this bound for an improper integral on an unbounded interval.


Examples

  • The series ∑n=0∞xn/n!\displaystyle\sum_{n=0}^\infty x^n/n! converges uniformly (by the Weierstrass MM-test) on every fixed interval [−R,R][-R,R], where R>0R>0: ∣xn/n!∣≤Rn/n!|x^n/n!|\le R^n/n!, and ∑nRn/n!=eR\displaystyle \sum_n R^n/n!=\mathrm{e}^R. It converges pointwise on R\mathbb{R}, but not uniformly there; for each finite partial sum, the difference from ex\mathrm{e}^x is unbounded as x→+∞x\to+\infty.
  • The geometric series ∑n=0∞xn\displaystyle \sum_{n=0}^\infty x^n converges uniformly on [−r,r][-r,r] for every fixed 0<r<10<r<1, because ∣xn∣≤rn|x^n|\le r^n and ∑nrn<∞\displaystyle \sum_n r^n<\infty. On (−1,1)(-1,1) it converges pointwise to 1/(1−x)1/(1-x), but not uniformly: the remainder xN+1/(1−x)x^{N+1}/(1-x) is unbounded as x→1−x\to1^-.

Prerequisites

Sources

Aliases: series of functions, function series Symbols: element of, infinity, integral, pi, real numbers, sum

Last updated: 20 October 2026 03:02