Power Series

Definition

1.A power series is a particular function series whose terms are the successive powers of the variable, each multiplied by a constant coefficient. It can be written in the form:

$$ \sum_{n=0}^{\infty} a_n x^n = a_0 + a_1 x + a_2 x^2 + \dots + a_n x^n + \dots$$

$$a_n \in \mathbb{R} \ \ \forall \ n \in \mathbb{N}, \ x \in \mathbb{R}$$

2.The real numbers $a_0, a_1, \dots, a_n, \dots$ are the coefficients of the series and $x$ is a real variable. By convention the symbol $a_0 x^0$ denotes $a_0$ for every $x$, including $x = 0,$ so that the term of index $0$ is always the constant $a_0.$ The $n$-th partial sum is the following polynomial of degree $n$:

$$ s_n(x) = a_0 + a_1 x + \dots + a_n x^n $$

3.An example is:

$$ \sum_{n=0}^{\infty} 3^n x^n = 1 + 3x + 9x^2 + 27x^3 + 81x^4 + \dots $$

4.whose coefficients are $a_n = 3^n$ and whose partial sums are the polynomials $s_n(x) = 1 + 3x + 9x^2 + \dots + 3^n x^n$.

5.The definition admits only non-negative integer powers of the variable, multiplied by coefficients that do not depend on $x$. Several expansions resemble a power series while violating one of these requirements.

6.An expansion that contains negative powers, is a Laurent series, not a power series. For example:

$$ \frac{1}{x} + a_0 + a_1 x + a_2 x^2 + \dots $$

7.When the exponents are fractional, as in $\sum_{n=0}^{\infty} a_n x^{n/2}$, the expansion is a Puiseux series.

8.An expression such as $\sum_{n=0}^{\infty}\cos(nx)x^n$, whose coefficients change with $x$, also fails the definition, because the coefficients must be constants.

9.A polynomial fits the definition as the special case in which the coefficients vanish from some index onward. A power series therefore extends a polynomial to infinitely many terms, and every polynomial is a power series that terminates.

Convergence and the theorem of Abel

10.Since power series are a special case of function series, the definitions and theorems of function series apply to them. Their rigid structure, however, makes the set of convergence much easier to describe. Two facts organise the theory:

12.The second statement is the theorem of Abel. Convergence at a single nonzero point therefore forces absolute convergence on the entire open interval up to that point, and divergence at a single point forces divergence everywhere beyond it.

13.Consider the series:

$$\sum_{n=0}^{\infty} \dfrac{x^n}{3^n}$$

14.At $x = 2$ it becomes a geometric series of ratio $2/3$, hence it converges. By Abel's theorem the series converges absolutely for $|x| < 2$, that is for $-2 < x < 2$. At $x = 6$ it becomes the geometric series $\sum_{n=0}^{\infty} 2^n$, which diverges, so by Abel's theorem the series cannot converge for $|x| > 6$ either.

Interval and radius of convergence

15.Let $E$ be the set of real $x$ at which the power series converges. This set always contains $x = 0$, so it is non-empty, and the radius of convergence is its supremum in absolute value:

$$ r = \sup\{\ |x| : x \in E \ \} $$

16.This number is the distance from the origin to the farthest point at which the series still converges, with $r = +\infty$ when $E$ is unbounded. By Abel's theorem $E$ is an interval symmetric about the origin, so the series converges for every $x \in (-r, r)$ and does not converge for $|x| > r$.

17.The interval $(-r, r)$ is the interval of convergence and $r$ is the radius of convergence.

19.The term radius becomes transparent once the variable is allowed to be complex. A power series $\sum_{n=0}^{\infty} a_n z^n$ in a complex variable converges absolutely on the open disk $|z| < r$, the disk of convergence, and diverges at every $z$ with $|z| > r$. More generally, a series centred at $z_0$ converges on the disk $|z - z_0| < r$. The real interval of convergence is the intersection of this disk with the real axis, so on the line we see the diameter of the disk while $r$ is its radius in the plane.

20.The behaviour at the endpoints $x = r$ and $x = -r$ is not determined by $r$ alone and must be examined separately in each case. To obtain the full set of convergence one finds the interval $(-r, r)$ and then studies the two endpoints individually.

The ratio criterion for the radius

21.The radius is encoded in the coefficients, so it is reasonable to expect a formula that recovers $r$ from them. The first such result follows from the ratio criterion of d'Alembert. Given $\sum_{n=0}^{\infty} a_n x^n$ with $a_n \neq 0$ from a certain index on, suppose the following limit exists, finite or infinite.

$$ \lim_{n \to +\infty} \left|\frac{a_{n+1}}{a_n}\right| = l $$

22.Then the radius of convergence is

$$ r = \begin{cases} \dfrac{1}{l} & l \neq 0 \\[8pt] +\infty & l = 0 \\[8pt] 0 & l = +\infty \end{cases} $$

23.The result is proved by applying the ratio criterion to the series of absolute values. The hypothesis gives the limit of consecutive ratios,

$$ \lim_{n \to +\infty} \left|\frac{a_{n+1} x^{n+1}}{a_n x^n}\right| = l |x| $$

24.When $l \neq 0$ the series converges for $l|x| < 1$, that is for $|x| < \tfrac{1}{l}$, which identifies the interval $\left(-\tfrac{1}{l}, \tfrac{1}{l}\right)$ and the radius $\tfrac{1}{l}$. When $l = 0$ the limit $l|x| = 0 < 1$ for every $x$, so the series converges on all of $\mathbb{R}$. When $l = +\infty$ and $x \neq 0$ the limit is $+\infty$, so convergence occurs only at $x = 0$.

25.For the series $\sum_{n=1}^{\infty} \dfrac{n^2 x^n}{3^n}$ the ratio criterion gives the limit:

$$ \begin{align} \lim_{n \to +\infty} \left|\frac{a_{n+1}}{a_n}\right| &= \lim_{n \to +\infty} \frac{(n+1)^2}{3^{\,n+1}} \cdot \frac{3^n}{n^2} \\[6pt] &= \frac{1}{3} \lim_{n \to +\infty} \left(\frac{n+1}{n}\right)^2 \\[6pt] &= \frac{1}{3} \lim_{n \to +\infty} \left(1+\frac{1}{n}\right)^2 \\[6pt] &= \frac{1}{3} \end{align} $$

26.So the radius of convergence is $r = 3$.

The root criterion for the radius

27.The second formula follows from the root criterion. Given $\sum_{n=0}^{\infty} a_n x^n$, suppose the following limit exists, finite or infinite:

$$ \lim_{n \to +\infty} \sqrt[n]{|a_n|} = l $$

28.Then the radius of convergence is

$$ r = \begin{cases} \dfrac{1}{l} & l \neq 0 \\[8pt] +\infty & l = 0 \\[8pt] 0 & l = +\infty \end{cases} $$

29.The proof applies the root criterion to:

$$\sqrt[n]{|a_n x^n|} = |x| \sqrt[n]{|a_n|}$$

30.The limit is $l|x|$, which leads to the same three cases as before. For example, for the series:

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

31.the root criterion gives the following limit:

$$ \lim_{n \to +\infty} \sqrt[n]{\frac{1}{(2n)^n}} = \lim_{n \to +\infty} \frac{1}{2n} = 0 $$

32.The radius of convergence is infinite and the series converges for every real $x$.

Power series centred at a point

33.Replacing $x$ by the difference $x - x_0$ shifts the centre of the series from the origin to the point $x_0$. A power series centred at $x_0 \in \mathbb{R}$ and $a_n \in \mathbb{R}$ is a function series of the form

$$ \sum_{n=0}^{\infty} a_n (x - x_0)^n = a_0 + a_1 (x - x_0) + a_2 (x - x_0)^2 + \dots $$

34.The case $x_0 = 0$ recovers a power series in $x$. The substitution $X = x - x_0$ turns the series into an ordinary power series in $X$, so every definition and property carries over. If $r$ is the radius of convergence, the series converges where $-r < x - x_0 < r$, which is the same as

$$ x_0 - r < x < x_0 + r $$

35.So the interval of convergence is the open interval $(x_0 - r, x_0 + r)$, centred at $x_0$ with radius $r$. Convergence at the endpoints $x_0 - r$ and $x_0 + r$ must be checked separately.

36.Consider the following series:

$$\sum_{n=1}^{\infty} \dfrac{(x+4)^n}{n}$$

37.It is centred at $x_0 = -4$. Setting $X = x + 4$ gives the power series:

$$\sum_{n=1}^{\infty} \dfrac{X^n}{n}$$

38.for which the ratio criterion yields

$$ \lim_{n \to +\infty} \left|\frac{a_{n+1}}{a_n}\right| = \lim_{n \to +\infty} \frac{n}{n+1} = 1 $$

39.so the radius is $1$ and the interval is $-1 < X < 1$. At $X = -1$ the series becomes the alternating harmonic series, which converges by the Leibniz criterion, while at $X = 1$ it becomes the harmonic series, which diverges. Returning to $x$, the original series converges for $-1 \leq x + 4 < 1$, that is $-5 \leq x < -3$.

Series reducible to power series

40.A series whose powers are taken of an expression in $x$, rather than of $x$ or of $x - x_0$, is not a power series, but a substitution may turn it into one.

41.Consider the series:

$$ \sum_{n=1}^{\infty} \frac{1}{n^2} (x^2 - 4)^n $$

42.It is not a power series because the base $x^2 - 4$ is neither $x$ nor a binomial $x - x_0$. Setting $X = x^2 - 4$ produces the power series $\sum_{n=1}^{\infty} \dfrac{X^n}{n^2}$, whose radius is $1$ since

$$ \lim_{n \to +\infty} \left|\frac{a_{n+1}}{a_n}\right| = \lim_{n \to +\infty} \frac{n^2}{(n+1)^2} = 1 $$

43.The series in $X$ converges for $-1 \leq X \leq 1$, because at $X = 1$ it is the generalized harmonic series of order $2$ and at $X = -1$ it converges by the Leibniz criterion. Translating back, $-1 \leq x^2 - 4 \leq 1$ gives $3 \leq x^2 \leq 5$, hence $-\sqrt{5} \leq x \leq -\sqrt{3}$ or $\sqrt{3} \leq x \leq \sqrt{5}$.

Algebraic operations on power series

44.Two power series with the same centre can be combined into new power series. Suppose the following series have radii of convergence $r_f$ and $r_g$:

$$ f(x) = \sum_{n=0}^{\infty} a_n x^n \qquad g(x) = \sum_{n=0}^{\infty} b_n x^n $$

45.Their sum and difference are obtained by adding or subtracting the coefficients of equal degree,

$$ f(x) \pm g(x) = \sum_{n=0}^{\infty} (a_n \pm b_n) x^n $$

46.The radius of this series is at least $\min(r_f, r_g)$, and it can be strictly larger when the dominant parts of the two coefficient sequences cancel.

47.The product is again a power series with the same centre, whose coefficients are the Cauchy products of the two sequences:

$$f(x)g(x) = \sum_{n=0}^{\infty} c_n x^n$$

$$c_n = \sum_{k=0}^{n} a_k b_{n-k}$$

48.Each $c_n$ collects every product $a_k b_{n-k}$ whose factors have degrees summing to $n$. This series converges at least on the smaller of the two intervals of convergence.

49.The quotient $f/g$ admits a power series expansion when $g(0) = b_0 \neq 0$. Writing $f/g = \sum_{n=0}^{\infty} d_n x^n$ and multiplying by $g$ turns the coefficients of $f$ into the Cauchy products of $(b_n)$ and $(d_n)$, so comparison of equal degrees gives

$$ a_n = \sum_{k=0}^{n} b_k d_{n-k} $$

50.Since $b_0 \neq 0$, this relation determines $d_0, d_1, d_2, \dots$ one degree at a time,

$$ d_n = \frac{1}{b_0}\left(a_n - \sum_{k=1}^{n} b_k d_{n-k}\right) $$

51.The increase of the radius for a sum appears with $a_n = 1$ and $b_n = 3^{-n} - 1$. Both series

$$ \sum_{n=0}^{\infty} x^n \qquad \sum_{n=0}^{\infty} (3^{-n} - 1) x^n $$

52.have radius $1$, since their coefficients approach constants of modulus $1$. Adding them gives coefficients $a_n + b_n = 3^{-n}$, hence the series $\sum_{n=0}^{\infty}\left(\frac{x}{3}\right)^n$, whose radius is $3$.

Uniform convergence of a power series

53.On any closed interval strictly inside the interval of convergence a power series converges totally, hence uniformly. A power series with radius $r > 0$ converges totally, and therefore uniformly, on every interval $[a, b]$ whose endpoints are interior points of the interval of convergence.

54.Consider the series:

$$\sum_{n=1}^{\infty} \dfrac{x^n}{n 2^n}$$

55.The ratio criterion gives interval of convergence $(-2, 2)$. Choosing a closed interval with interior endpoints, for example $[-0.5, 1]$, every $x$ in it satisfies:

$$ \left|\frac{x^n}{n 2^n}\right| \leq \frac{1}{n 2^n} \qquad \forall \ n \in \mathbb{N} $$

56.Since the series $\sum_{n=1}^{\infty} \dfrac{1}{n2^n}$ converges, the Weierstrass criterion gives uniform convergence on $[-0.5, 1]$. On every closed and bounded interval contained in the interval of convergence a power series therefore enjoys the properties of uniformly convergent series, including continuity of the sum and differentiation and integration term by term.

Term-by-term differentiation

57.The derived series of $\sum_{n=0}^{\infty} a_n x^n$ is the power series obtained by differentiating each term,

$$ \sum_{n=1}^{\infty} n a_n x^{n-1} $$

58.The derived series is again a power series, and it has the same interval of convergence as the original one. Since the terms of a power series are monomials, hence continuous and differentiable, the theorems on continuity and term-by-term differentiation apply, giving the following statement. If $\sum_{n=0}^{\infty} a_n x^n$ has interval of convergence $(-r, r)$ with $r > 0$ and sum $f(x)$, then $f$ is continuous and differentiable at every point of $(-r, r)$, and

$$ f'(x) = \sum_{n=1}^{\infty} n a_n x^{n-1} = a_1 + 2 a_2 x + 3 a_3 x^2 + \dots $$

59.Because the derived series is itself a power series with the same radius, the argument can be repeated, so the sum of a power series is differentiable infinitely many times on the open interval of convergence, and every derivative is obtained by differentiating the series term by term.

60.The geometric series:

$$\sum_{n=0}^{\infty} x^n$$

61.has radius $1$ and sum $f(x) = \dfrac{1}{1-x}$ on $(-1, 1)$. Differentiating the sum gives:

$$f'(x) = \dfrac{1}{(1-x)^2}$$

62.Differentiating the series term by term gives $\sum_{n=1}^{\infty} n x^{n-1}$, so:

$$ \frac{1}{(1-x)^2} = \sum_{n=1}^{\infty} n x^{n-1} \qquad x \in (-1, 1) $$

63.A further differentiation yields

$$ \frac{2}{(1-x)^3} = \sum_{n=2}^{\infty} n(n-1) x^{n-2} \qquad x \in (-1, 1) $$

64.and the process continues indefinitely.

Term-by-term integration

65.The argument that justifies differentiating a power series term by term applies equally to integration. The integrated series of $\sum_{n=0}^{\infty} a_n x^n$ is obtained by integrating each term from $0$ to $x$:

$$ \sum_{n=0}^{\infty} \frac{a_n}{n+1} x^{n+1} $$

66.This is again a power series, and it has the same radius of convergence as the original one. If $\sum_{n=0}^{\infty} a_n x^n$ has interval of convergence $(-r, r)$ with $r > 0$ and sum $f(x)$, then $f$ is integrable on every closed interval contained in $(-r, r)$, and the following holds:

$$ \int_0^x f(t) \ dt = \sum_{n=0}^{\infty} \frac{a_n}{n+1} x^{n+1} \qquad x \in (-r, r) $$

67.The constant of integration is fixed by the value at $x = 0$, where both sides vanish.

68.The geometric series $\sum_{n=0}^{\infty} x^n$ has sum $\dfrac{1}{1-x}$ on $(-1, 1)$. Integrating both representations from $0$ to $x$ gives, on the left,

$$ \int_0^x \frac{dt}{1-t} = -\ln(1-x) $$

69.and on the right the integrated series $\sum_{n=0}^{\infty} \dfrac{x^{n+1}}{n+1}$. Shifting the summation index produces the expansion

$$ -\ln(1-x) = \sum_{n=1}^{\infty} \frac{x^n}{n} \qquad x \in (-1, 1) $$

70.which represents the logarithm as a power series throughout the interval of convergence.