Function Series

Sequences of functions

1.Take a set $A \subseteq \mathbb{R}$ and a rule that attaches a function $f_n$ to every natural number $n$. The result is no longer a list of numbers but a list of functions that all share the same domain $A$:

$$ f_0(x), \ f_1(x), \ f_2(x), \ \dots, \ f_n(x), \ \dots $$

2.An object of this kind is a sequence of functions on $A$. The next family shows the idea at work:

$$ f_n(x) = \frac{x^n}{1+x^n} \qquad x \in [0, +\infty) \tag{1} $$

3.For a fixed exponent $n$ it is a single function of $x$, and as $n$ ranges over $\mathbb{N}$ it produces infinitely many of them. On the chosen domain the quantity $1+x^n$ stays positive, so no term has a gap.

Pointwise convergence of a sequence of functions

4.Freeze $x$ at one value and the functions $f_n(x)$ reduce to plain numbers, one per index $n$. Their limit measures what the family does at that single location, and sweeping the value across the domain assembles the global picture from these local ones, treating each $f_n(x)$ as a numerical sequence.

5.Applied to $(1)$, the analysis produces three regimes according to the size of $x:$

7.Collecting the three outcomes, the limit is the function:

$$ f(x) = \begin{cases} 0 & 0 \leq x < 1 \\[6pt] \dfrac{1}{2} & x = 1 \\[6pt] 1 & x > 1 \end{cases} $$

8.Each $f_n$ is continuous, yet the limit breaks at $x = 1$. Continuity is therefore not handed down automatically by this mode of convergence, an observation that prepares the ground for the stronger mode introduced further on.

9.We say that $f_n$, defined on the common domain $A \subseteq \mathbb{R}$, converges pointwise on a subset $E \subseteq A$ if at each $x \in E$ the numbers $f_n(x)$ tend to a finite limit. Computed point by point, these limits define a function on $E$:

$$ f(x) = \lim_{n \to +\infty} f_n(x) \qquad \forall x \in E $$

10.The largest such $E$ is the set of pointwise convergence. At a point where $f_n(c)$ fails to settle on a finite value, the family is divergent or indeterminate there.

Function series

11.Starting from a sequence of functions $f_n$ on $A \subseteq \mathbb{R}$, build the running totals in the same manner used for numerical series:

$$ \begin{align} s_0(x) &= f_0(x) \\[6pt] s_1(x) &= f_0(x) + f_1(x) \\[6pt] s_2(x) &= f_0(x) + f_1(x) + f_2(x) \\[6pt] &\ \vdots \\[6pt] s_n(x) &= f_0(x) + f_1(x) + \dots + f_n(x) \end{align} $$

12.Pushing $n$ to infinity turns these totals into a sum of infinitely many functions, the function series generated by $f_n$,

$$ \sum_{n=0}^{\infty} f_n(x) $$

13.Within this object $f_n(x)$ is the general term and $s_n(x)$ is the partial sum of order $n$. The remainder of order $n$, written $r_n(x)$, collects the terms past the first $n+1$:

$$ r_n(x) = \sum_{k=n+1}^{\infty} f_k(x) = f_{n+1}(x) + f_{n+2}(x) + \dots $$

14.The shared domain $A$ becomes the domain of the series. Among all function series, two types are studied in their own right, the power series with monomial terms and the Fourier series with trigonometric terms.

15.

Replacing $x$ by a number turns the series into a numerical one. With $\sum_{n=0}^{\infty} x^n$ this substitution gives a geometric series whose ratio is exactly the number inserted.

Pointwise convergence of a function series

16.At a fixed point the series behaves like the numerical series obtained by evaluating each term there. Writing $\sum_{n=0}^{\infty} f_n(x)$ for $x \in A \subseteq \mathbb{R}$, the character at a point $c$ is read off from $\sum_{n=0}^{\infty} f_n(c)$:

18.Gathering every $x$ at which convergence holds produces the set of pointwise convergence $E \subseteq A$. On $E$ the series defines a sum function $s : E \to \mathbb{R}$:

$$ s(x) = \sum_{n=0}^{\infty} f_n(x) \qquad \forall x \in E $$

19.The series then converges pointwise to $s(x)$ on $E$. Locating $E$ amounts to holding $x$ as a parameter and falling back on the criteria for numerical series.

20.A worked case is the following series:

$$ \sum_{n=1}^{\infty} \frac{1}{(1+x^2)^n} $$

21.For fixed $x$ this is geometric with ratio $\dfrac{1}{1+x^2}$. That ratio lies in $(0, 1]$ for every real $x$ and reaches $1$ only at the origin, so convergence holds at every point except $x = 0$. The set of pointwise convergence is

$$ E = \{\ x \in \mathbb{R} \mid x \neq 0 \ \} $$

22.On $E$ the geometric sum formula gives the closed form:

$$ s(x) = \frac{\dfrac{1}{1+x^2}}{1-\dfrac{1}{1+x^2}} = \frac{1}{x^2} $$

23.At each nonzero point the partial sums settle on this value.

Uniform convergence

24.In pointwise convergence the index that pushes the remainder $r_n(x)$ below a prescribed $\varepsilon$ may vary from point to point, and nothing stops it from growing without bound as $x$ moves. Requiring instead one index valid at every point at once gives a stronger property.

25.The series $\sum_{n=0}^{\infty} f_n(x)$, convergent to $s(x)$ on $E \subseteq \mathbb{R}$, converges uniformly to $s(x)$ on a subset $F \subseteq E$ when

$$ \forall \varepsilon > 0 \ \ \exists p \in \mathbb{N} \ : \ \forall n > p \ \text{and} \ \forall x \in F, \quad |r_n(x)| < \varepsilon $$

26.Uniform convergence on a set forces pointwise convergence on it, so the uniform-convergence set sits inside the pointwise one and can be strictly smaller.

27.The geometric series $\sum_{n=0}^{\infty} x^n$ makes the gap visible. When $|x| < 1$ its remainder closes in geometric form,

$$ r_n(x) = \sum_{k=n+1}^{\infty} x^k = \frac{x^{n+1}}{1-x} $$

28.Confine $x$ to $[-\rho, \rho]$ with $0 \leq \rho < 1$: the numerator is at most $\rho^{n+1}$ and the denominator at least $1-\rho$, which yields

$$ |r_n(x)| = \frac{|x|^{n+1}}{|1-x|} \leq \frac{\rho^{n+1}}{1-\rho} \qquad \forall \ x \in [-\rho, \rho] $$

29.This estimate ignores $x$ and vanishes as $n$ grows, so one threshold serves the whole interval and convergence is uniform on $[-\rho, \rho]$. Over the open interval $]-1, 1[$ the picture breaks down, since letting $x \to 1^-$ drives $\dfrac{x^{n+1}}{1-x}$ to infinity for every fixed $n$, the remainder admits no uniform bound, and uniform convergence fails although pointwise convergence persists.

The Cauchy criterion for uniform convergence

30.A test that bypasses the sum altogether works by keeping finite stretches of consecutive terms small. The series $\sum_{n=0}^{\infty} f_n(x)$ converges uniformly on $E \subseteq \mathbb{R}$ exactly when, given $\varepsilon > 0$, some index $n_\varepsilon \in \mathbb{N}$ makes the following expression hold for all $n > m > n_\varepsilon$ and all $x \in E$:

$$ |f_{m+1}(x) + f_{m+2}(x) + \dots + f_n(x)| < \varepsilon $$

31.This mirrors the Cauchy criterion for numerical series, with the decisive feature that $n_\varepsilon$ is fixed by $\varepsilon$ alone and never tuned to $x$.

Total convergence and the Weierstrass criterion

32.Direct estimates on the remainder are frequently out of reach, which makes a comparison against a numerical series attractive. Suppose that for $\sum_{n=0}^{\infty} f_n(x)$ on $E \subseteq \mathbb{R}$ there are nonnegative constants $M_n$, forming a convergent series $\sum_{n=0}^{\infty} M_n$, that bound every term:

$$ |f_n(x)| \leq M_n \qquad \forall \ n \in \mathbb{N}, \ \ \forall \ x \in E $$

33.Then convergence is uniform on $E$. The majorants pin each tail of the function series beneath the matching tail of $\sum_{n=0}^{\infty} M_n$, already uniformly small by the Cauchy condition.

34.On the whole real line the series $\sum_{n=1}^{\infty} \dfrac{\cos(nx)}{n^2}$ fits the scheme. Because the cosine never exceeds $1$ in absolute value, each term carries a constant majorant:

$$ \left|\frac{\cos(nx)}{n^2}\right| \leq \frac{1}{n^2} \quad \forall \ n \in \mathbb{N}, \ \forall \ x \in \mathbb{R} $$

35.The series $\sum_{n=1}^{\infty} \dfrac{1}{n^2}$ converges, so the hypotheses hold and the series converges uniformly on $\mathbb{R}$.

36.

A series admitting such majorants is called totally convergent. Total convergence yields uniform convergence, and because $\sum_{n} |f_n(x)| \leq \sum_{n} M_n$, it yields absolute and therefore pointwise convergence as well.

Passage to the limit term by term

37.The results that follow all express one principle, under uniform convergence an operation can be moved inside the infinite sum. Limits come first. Suppose $\sum_{n=0}^{\infty} f_n(x)$ converges uniformly on $E \subseteq \mathbb{R}$ to $s(x)$, let $c$ be an accumulation point of $E$, and assume every term has a finite limit:

$$ \lim_{x \to c} f_n(x) = a_n \qquad \forall n \in \mathbb{N} $$

38.Then $\sum_{n=0}^{\infty} a_n$ converges and the two limiting processes commute:

$$ \lim_{x \to c} \sum_{n=0}^{\infty} f_n(x) = \sum_{n=0}^{\infty} \lim_{x \to c} f_n(x) $$

39.This computes the limit of the sum even when no formula for the sum is available. Take $\sum_{n=1}^{\infty} \dfrac{1}{n^2+x}$ on $[0, +\infty[$. The inequality $\dfrac{1}{n^2+x} \leq \dfrac{1}{n^2}$ secures uniform convergence, while $\dfrac{1}{n^2+x} \to \dfrac{1}{n^2}$ as $x \to 0$, so the limit may pass through the sum:

$$ \lim_{x \to 0} \sum_{n=1}^{\infty} \frac{1}{n^2+x} = \sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6} $$

Continuity of the sum

40.Continuity survives the passage to a uniformly convergent sum. Whenever $\sum_{n=0}^{\infty} f_n(x)$ converges uniformly on $E \subseteq \mathbb{R}$ and each term is continuous at a point $c \in E$, the sum $s(x)$ is continuous at $c$, and continuity of the terms throughout $E$ makes $s(x)$ continuous on all of $E$. The proof selects a partial sum lying uniformly near $s$ and transfers the continuity of that partial sum to $s$ in a neighbourhood of $c$.

41.Consider $\sum_{n=1}^{\infty} \dfrac{\cos^n x}{2^n}$ on $\mathbb{R}$.

42.The majorant $\left|\dfrac{\cos^n x}{2^n}\right| \leq \dfrac{1}{2^n}$ paired with the convergent $\sum_{n=1}^{\infty} \dfrac{1}{2^n}$ delivers uniform convergence by the Weierstrass criterion, and since every term is continuous the sum is continuous too. Here the sum is a geometric series of ratio $\tfrac{1}{2}\cos x$ starting at $n = 1$, with closed form:

$$ s(x) = \frac{\tfrac{1}{2}\cos x}{1-\tfrac{1}{2}\cos x} = \frac{\cos x}{2-\cos x} $$

43.This function is continuous on all of $\mathbb{R}$.

Integration term by term

44.For finitely many functions the integral distributes over the sum, and uniform convergence extends this to infinite sums. If $\sum_{n=0}^{\infty} f_n(x)$ converges uniformly on $[a, b]$ with each term integrable on $[a, b]$, then $s(x)$ is integrable on $[a, b]$ and

$$ \int_a^b \left[\sum_{n=0}^{\infty} f_n(x)\right] dx = \sum_{n=0}^{\infty} \int_a^b f_n(x) \ dx $$

45.The exchange runs both ways, returning the integral of the sum from the integrated terms, or the sum of a numerical series from those integrals.

46.On $\left[0, \tfrac{1}{2}\right]$ the geometric series converges uniformly and its powers are integrable. Integrating each term gives a chain of equalities:

$$ \int_0^{1/2} \frac{1}{1-x} \ dx = \sum_{n=0}^{\infty} \int_0^{1/2} x^n \ dx = \sum_{n=0}^{\infty} \frac{1}{(n+1)2^{n+1}} $$

47.The integral on the left equals $\ln 2$, so shifting the summation index pins down a numerical series:

$$ \sum_{n=1}^{\infty} \frac{1}{n2^n} = \ln 2 $$

48.This value is reached without ever summing the original series of functions.

Differentiation term by term

49.Differentiation is more delicate, because a derivative can magnify oscillations that the functions themselves keep small, so uniform convergence of the series by itself does not suffice. The condition that works is uniform convergence of the differentiated series.

50.Let $\sum_{n=0}^{\infty} f_n(x)$ converge on $[a, b]$ with each term differentiable on $[a, b]$, and let $\sum_{n=0}^{\infty} f_n'(x)$ converge uniformly on $[a, b]$. Then $s(x)$ is differentiable on $[a, b]$, and its derivative is obtained term by term:

$$ s'(x) = \sum_{n=0}^{\infty} f_n'(x) \qquad \forall x \in [a, b] $$

51.The derivative of the sum coincides with the sum of the derivatives. Behind the extra hypothesis lies the mean value theorem, applied to the differences $f_m - f_n$ to connect increments of the partial sums with their derivatives.

52.Differentiating the geometric series produces a new closed form. Fix an interval $[-\rho, \rho]$ with $0 \leq \rho < 1$. The differentiated terms satisfy $|nx^{n-1}| \leq n\rho^{n-1}$, and $\sum_{n=1}^{\infty} n\rho^{n-1}$ converges, so $\sum_{n=1}^{\infty} nx^{n-1}$ converges uniformly on that interval.

53.Differentiating $\sum_{n=0}^{\infty} x^n = \dfrac{1}{1-x}$ term by term then yields a closed form:

$$ \sum_{n=1}^{\infty} n x^{n-1} = \frac{d}{dx}\left(\frac{1}{1-x}\right) = \frac{1}{(1-x)^2} \qquad -1 < x < 1 $$

54.The identity holds throughout $(-1, 1)$, since every point of it lies in some interval $[-\rho, \rho]$.