Infinite Series of Functions¶

Definition¶

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

Interpretation¶

A series of functions is defined through its partial sums: each $S_N$ is an ordinary finite sum of functions, and the series converges to $f$ when the sequence $(S_N)$ 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 $1$ , $2$ , $4$ and $10$ terms of $\displaystyle \sum_{n\ge1}\sin(nx)/n^2$ on $[0,2\pi]$ . Each new term changes the curve less, and the later sums are almost indistinguishable. Since $|\sin(nx)/n^2|\le1/n^2$ for every $x$ , the sum with $10$ terms differs from the full sum by at most $\displaystyle \sum_{n>10}1/n^2\approx0.095$ at every point at once; this is the bound used in the Weierstrass $M$ -test.

Raise $N$ : the partial sums $\displaystyle S_N(x)=\sum_{n=1}^N f_n(x)$ settle on the limit function (dashed).

Pointwise convergence at $x$ is convergence of the numerical infinite series $\displaystyle \sum_n f_n(x)$ , so the tests for numerical series apply for each fixed $x$ . Uniform convergence asks more: the largest error $\sup_{x\in X}|f(x)-S_N(x)|$ must tend to $0$ . 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: $\displaystyle \sum_{n\ge0}x^2/(1+x^2)^n$ equals $1+x^2$ for $x\ne0$ and $0$ at $x=0$ $0$ .

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: $\displaystyle \sum_n\sin(nx)/n^2$ converges uniformly, but its derivative series $\displaystyle \sum_n\cos(nx)/n$ is the harmonic series at $x=0$ and diverges.

Key Results¶

If $f_n$ 's are (continuously) differentiable, $\displaystyle \sum_{n=1}^\infty f_n \rightarrow f\text{,}$ and $\displaystyle \sum_{n=1}^\infty f'_n\rightrightarrows g$ on $(a,b)\text{,}$ then $f$ is (continuously) differentiable on $(a, b)$ with $f' = g\text{,}$ i.e., for every $x_0\in(a,b)$ $(a, b)$ ,

\sum_{n=1}^\infty f'_n(x_0)=\left(\sum_{n=1}^\infty f_n\right)'(x_0).

On a finite closed interval $[a,b]$ with $a<b$ $b$ , if every $f_n$ is Riemann integrable and $\displaystyle \sum_{n=1}^\infty f_n$ converges uniformly, then its sum is Riemann integrable and

\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 $(b-a)$ times that error. Uniform convergence alone does not provide this bound for an improper integral on an unbounded interval.

Examples¶

The series $\displaystyle\sum_{n=0}^\infty x^n/n!$ converges uniformly (by the Weierstrass $M$ -test) on every fixed interval $[-R,R]$ , where $R>0$ $0$ : $|x^n/n!|\le R^n/n!$ $R^{n} / n!$ , and $\displaystyle \sum_n R^n/n!=\mathrm{e}^R$ $e^{R}$ . It converges pointwise on $\mathbb{R}$ , but not uniformly there; for each finite partial sum, the difference from $\mathrm{e}^x$ is unbounded as $x\to+\infty$ $+ \infty$ .
The geometric series $\displaystyle \sum_{n=0}^\infty x^n$ converges uniformly on $[-r,r]$ for every fixed $0<r<1$ $1$ , because $|x^n|\le r^n$ and $\displaystyle \sum_n r^n<\infty$ $\infty$ . On $(-1,1)$ it converges pointwise to $1/(1-x)$ $x)$ , but not uniformly: the remainder $x^{N+1}/(1-x)$ is unbounded as $x\to1^-$ $1^{-}$ .

Prerequisites¶

Sources¶

Math III: Analysis · Lecture 2

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

Last updated: 20 October 2026 03:02