5.11. Convergent Power Series

This chapter is up to a certain generalization of the polynomial idea. As a polynomial is the finit sum of its monoms, it is advisable to consider infinite sums of this kind, i.e. we are going to study series of functions. Of course this is just a formal concept at this stage. Whether it results in a genuine function, capable to accept real numbers, surely depends on convergence properties. It is an advantage that polynomials calculate only with addition and multiplication. Although this simplicity certainly will get lost with our prospective generalized polynomials the original advantage will apply again if we are only interested in approximative values.

Definition: For every sequence

{(a_{n})}_{n \geq 0}

ℝ

and every

a \in ℝ

the series of functions

(\sum_{i = 0}^{n} a_{i} {(X - a)}^{i})

[5.11.1]

is called a power series with a as expansion point.

For every $x \in ℝ$ the power series $(\sum_{i = 0}^{n} a_{i} {(X - a)}^{i})$ provides a series of numbers by evaluating X at x:

(\sum_{i = 0}^{n} a_{i} {(x - a)}^{i})

This series is always convergent for the expansion point a: $(\sum_{i = 0}^{n} a_{i} {(a - a)}^{i}) = (a_{0})_{n \geq 0}$ . Thus we will focus on those power series which converge for at least one further number.

Definition: We call a power series $(\sum_{i = 0}^{n} a_{i} {(X - a)}^{i})$ convergent, if it converges for at least two different points, i.e. if there is an $y \neq a$ such that

(\sum_{i = 0}^{n} a_{i} {(y - a)}^{i})

converges.

[5.11.2]

Power series which are convergent for their expansion point solely are called divergent.

Surprisingly the convergence for only one additional point leads to the convergence for a whole neighbourhood of a.

Proposition: If $(\sum_{i = 0}^{n} a_{i} {(X - a)}^{i})$ converges in a $y \neq a$ it converges (even absolutly) for every

x \in] a - | y - a |, a + | y - a | [

[5.11.3]

Proof: As the series $(\sum_{i = 0}^{n} a_{i} {(y - a)}^{i})$ converges, the sequence $(a_{n} {(y - a)}^{n})_{n \geq 0}$ is a zero sequence according to [5.9.9], thus bounded. So there is a c such that

| a_{i} {(y - a)}^{i} | \leq c \Leftrightarrow | a_{i} | \leq \frac{c}{| y - a |^{i}} for all i \in ℕ

Thus the following estimate holds for all i:

| a_{i} {(x - a)}^{i} | = | a_{i} | | x - a |^{i} \leq c \cdot | \frac{x - a}{y - a} |^{i}

From $| x - a | < | y - a |$ we have $| \frac{x - a}{y - a} | < 1$ . Consequently the geometric series $(\sum_{i = 0}^{n} c \cdot | \frac{x - a}{y - a} |^{i})$ converges and dominates the series $(\sum_{i = 0}^{n} a_{i} {(x - a)}^{i})$ which is convergent now due to the comparison test [5.9.15].

So we know that a convergent power series never converges for isolated points only. In fact its domain of convergence will always be an open interval, which allows to introduce the following notions.

Definition: For any power series $(\sum_{i = 0}^{n} a_{i} {(X - a)}^{i})$ the number

r ≔ \sup {| y - a | | y \in ℝ \land (\sum_{i = 0}^{n} a_{i} {(y - a)}^{i}) converges}

[5.11.4]

is called its radius of convergence (we liberally allow $r = \infty$ ^*) as a value). If $r > 0$ we call the open interval

] a - r, a + r [

[5.11.5]

the domain of convergence.

___________
^*) We set: $\infty > 0$ and $\infty \pm c = \infty$ for every real number c.

Consider:

$r = 0 \Leftrightarrow (\sum_{i = 0}^{n} a_{i} {(X - a)}^{i})$ is divergent.
$r = \infty \Leftrightarrow (\sum_{i = 0}^{n} a_{i} {(X - a)}^{i})$ the domain of convergence is the whole of $ℝ$ .
$| x - a | > r \Rightarrow (\sum_{i = 0}^{n} a_{i} {(x - a)}^{i})$ is divergent.

Calculating a radius of convergence according to [5.11.4] is an akward task, so there is a need for other techniques.

Proposition: Let $(\sum_{i = 0}^{n} a_{i} {(X - a)}^{i})$ be a power series and r its radius of convergence. If $a_{i} \neq 0$ for all the coefficients, we have:

$\| \frac{a_{n + 1}}{a_{n}} \| \to 0 \Rightarrow r = \infty$	[5.11.6]
$(\| \frac{a_{n}}{a_{n + 1}} \|)_{n \geq 0} convergent \Rightarrow r = \lim \| \frac{a_{n}}{a_{n + 1}} \|$	[5.11.7]

Proof: We employ the ratio test [5.9.16] several times. Note that if $x \neq a$ we may calculate

| \frac{a_{i + 1} {(x - a)}^{i + 1}}{a_{i} {(x - a)}^{i}} | = | x - a | \cdot | \frac{a_{i + 1}}{a_{i}} |

[0]

1. ►

From $\frac{a_{n + 1}}{a_{n}} \to 0$ we see that for each x the sequence $(| x - a | \cdot | \frac{a_{n + 1}}{a_{n}} |)_{n \geq 0}$ is a zero sequence as well. So there is a number $k \in ℕ$ such that $| x - a | \cdot | \frac{a_{i + 1}}{a_{i}} | < \frac{1}{2} for all i \geq k$ . Due to the ratio test the power series thus converges for every $x \neq a$ .

2. ►

We set as an abbreviation $r^{'} ≔ \lim | \frac{a_{n}}{a_{n + 1}} | \geq 0$ and prove:

$r^{'} \geq r$ : If $| x - a | > r^{'}$ we find a number $k \in ℕ$ such that (note: $x \neq a$ )

$| \frac{{(x - a)}^{i + 1}}{{(x - a)}^{i}} | = | x - a | > | \frac{a_{i}}{a_{i + 1}} |$

for all $i \geq k$ . Hence we have for these i: $| a_{i + 1} {(x - a)}^{i + 1} | \geq | a_{i} {(x - a)}^{i} |$ and thus (by an easy induction argument)

$| a_{i} {(x - a)}^{i} | \geq | a_{k} {(x - a)}^{k} | > 0$ .

So $(a_{n} {(x - a)}^{n})_{n \geq 0}$ is no zero sequence and consequently the series $(\sum_{i = 0}^{n} a_{i} {(x - a)}^{i})$ is divergent. Now if $r^{'} < r$ , we would find an x with $| x - a | > r^{'}$ such that the power series is convergent for, which contradicts the result above.
$r^{'} \leq r$ : If $| x - a | < r^{'}$ we choose two numbers $c_{1}, c_{2}$ such that $| x - a | < c_{1} < c_{2} < r^{'}$ . From the limit properties of $r^{'}$ we get a $k \in ℕ$ such that

$c_{2} < | \frac{a_{i}}{a_{i + 1}} |$

for all $i \geq k$ . These i allow to estimate as follows:

$| x - a | \cdot | \frac{a_{i + 1}}{a_{i}} | = \frac{| x - a |}{| \frac{a_{i}}{a_{i + 1}} |} < \frac{c_{1}}{c_{2}} < 1$ .

[0] and the ratio test hence guarantee the convergence of the power series for $x \neq a$ . Now $r^{'} > r$ can't be true as otherwise we could choose an x with $r < | x - a | < r^{'}$ , i.e. an x outside the domain of convergence, the series will converge for.

Consider:

As convergence is not affected by the first members the above criterion is also valid if the condition $a_{i} \neq 0$ is only true for almost all i.

Another way for calculating r is the Hadamard formula, an implication of the root test [5.9.17]. We note this formula without proof:

If $(\sqrt[n]{| a_{n} |})_{n \geq 0}$ is bounded there is a greatest accumulation point, the so called supremum limit (symbolically $\lim^{¯} \sqrt[n]{| a_{n} |}$ ). The radius of convergence now calculates to:

r = {\begin{array}{l} \frac{1}{\lim^{¯} \sqrt[n]{| a_{n} |}},   if \lim^{¯} \sqrt[n]{| a_{n} |} \neq 0 \\ \infty,   if \lim^{¯} \sqrt[n]{| a_{n} |} = 0 \end{array}

[5.11.8]

If $(\sqrt[n]{| a_{n} |})_{n \geq 0}$ is unbounded we have $r = 0$ .

Proposition and Notation: Every convergent power series $(\sum_{i = 0}^{n} a_{i} {(X - a)}^{i})$ converges (absolutly) for each point x in its domain of convergence. The function

f :] a - r, a + r [\to ℝ, f (x) ≔ \sum_{i = 0}^{\infty} a_{i} {(x - a)}^{i}

[5.11.9]

is called its limit function which we denote by the symbol $\sum_{i = 0}^{\infty} a_{i} {(X - a)}^{i}$ .

Proof: If $x \in] a - r, a + r [$ , i.e. $| x - a | < r$ , the number $| x - a |$ must be exceeded by an element of that set which has r as supremum. Hence $(\sum_{i = 0}^{n} a_{i} {(X - a)}^{i})$ will converge for at least one y such that

| x - a | < | y - a |

According to [5.11.3] $(\sum_{i = 0}^{n} a_{i} {(X - a)}^{i})$ converges absolutely for x.

Consider:

Convergence behaviour at the bounderies of the domain of convergence varies. For example: Employing [5.11.7] we calculate the radius of convergence for the power series $(\sum_{i = 0}^{n} \frac{1}{i + 1} X^{i})$ (with expansion point 0) to

$r = \lim \frac{\frac{1}{n + 1}}{\frac{1}{n + 2}} = \lim \frac{n + 2}{n + 1} = 1$ .

$(\sum_{i = 0}^{n} \frac{1}{i + 1} X^{i})$ converges for $- 1$ (alternating harmonic series [5.9.12]), but not for 1 (harmonic series [5.9.6]).

If $| x | < r$ the number $x + a$ belongs to the domain of convergence $] a - r, a + r [$ . Consequently the series

(\sum_{i = 0}^{n} a_{i} x^{i}) = (\sum_{i = 0}^{n} a_{i} {(x + a - a)}^{i})

[5.11.10]

converges absolutely.

Example:

$(\sum_{i = 0}^{n} {(- 1)}^{i} {(X - 1)}^{i})$ is a power series expanded at 1. Due to [5.11.7] the radius of convergence is 1 which results in $] 0,2 [$ as domain of convergence. Considering the limit of the geometric series ([5.9.4]) the limit function turns out to be the restriction of the reciprocal function to $] 0,2 [$ : For any $x \in] 0,2 [$ we have a distance $| 1 - x | < 1$ , thus:

\sum_{i = 0}^{\infty} {(- 1)}^{i} {(x - 1)}^{i} = \sum_{i = 0}^{\infty} {(1 - x)}^{i} = \frac{1}{1 - (1 - x)} = \frac{1}{x}

[5.11.11]

Each of the power series

(\sum_{i = 0}^{n} \frac{1}{i!} X^{i})

(\sum_{i = 0}^{n} \frac{{(- 1)}^{i}}{(2 i + 1)!} X^{2 i + 1})

(\sum_{i = 0}^{n} \frac{{(- 1)}^{i}}{(2 i)!} X^{2 i})

[5.11.12]

expanded at 0 converges for the whole of $ℝ$ according to [5.9.18], [5.9.19], [5.9.20]. Their limit functions are exp, sin und cos respectively.

Let $p = \sum_{i = 0}^{k} a_{i} X^{i}$ be an arbitrary polynomial. If we define $a_{i} ≔ 0, i > k$ , the set of coefficients $a_{0}, \dots, a_{k}$ will be extanded to a sequence such that the power series $(\sum_{i = 0}^{n} a_{i} X^{i})$ converges for the whole of $ℝ$ . The limit function

\sum_{i = 0}^{\infty} a_{i} X^{i} = \sum_{i = 0}^{k} a_{i} X^{i}

[5.11.13]

equals p. So the polynomials are the limit functions of certain power series!

Now we concentrate on the properties of convergent power series. The first idea is to adequately transfer the identity theorem for polynomials (cf. [4.5.*]) to convergent power series: A limit function already equals zero if it has countable many (suitable) zeros!

Proposition (Identity Theorem for Convergent Power Series): Let $(\sum_{i = 0}^{n} a_{i} {(X - a)}^{i})$ be convergent and $(x_{n})$ a sequence in $] a - r, a + r [$ such that $a \neq x_{n} \to a$ . If all sequence members are zeros of the limit function we have:

a_{k} = 0 for all k \in ℕ

[5.11.14]

Proof: As $x_{n} \to a$ we have a positive number $s < r$ such that $x_{n} \in [a - s, a + s] for all n \in ℕ^{*}$ . The limit $c ≔ \sum_{i = 0}^{\infty} | a_{i} | s^{i}$ exists due to [5.11.10] so that we may estimate for all x with $| x - a | \leq s$ and for all $j \in ℕ$ like this:

| \sum_{i = j}^{\infty} a_{i} {(x - a)}^{i - j} | \leq \sum_{i = j}^{\infty} | a_{i} | s^{i - j} = \frac{1}{s^{j}} \sum_{i = j}^{\infty} | a_{i} | s^{i} \leq \frac{c}{s^{j}}

[1]

We now prove [5.11.14] by induction:

As all sequence members are zeros of the limit function we get with $j = 1$ in [1] for all n:
$\begin{array}{l} 0 & = \sum_{i = 0}^{\infty} a_{i} {(x_{n} - a)}^{i} \\ = a_{0} + (x_{n} - a) \sum_{i = 1}^{\infty} a_{i} {(x_{n} - a)}^{i - 1} \\ \Rightarrow & | a_{0} | & = | (x_{n} - a) \sum_{i = 1}^{\infty} a_{i} {(x_{n} - a)}^{i - 1} | \\ \leq | x_{n} - a | \cdot \frac{c}{s} \end{array}$
Thus the convergence $x_{n} \to a$ results in $a_{0} = 0$ .
Let $a_{0} = \dots = a_{k} = 0$ be true. We employ [1] with $j = k + 2$ and hence get for all n:
$\begin{array}{l} 0 & = \sum_{i = k + 1}^{\infty} a_{i} {(x_{n} - a)}^{i} \\ = \underset{\neq 0}{\underset{︸}{{(x_{n} - a)}^{k + 1}}} \sum_{i = k + 1}^{\infty} a_{i} {(x_{n} - a)}^{i - k - 1} \\ \Rightarrow & 0 & = \sum_{i = k + 1}^{\infty} a_{i} {(x_{n} - a)}^{i - k - 1} \\ = a_{k + 1} + (x_{n} - a) \sum_{i = k + 2}^{\infty} a_{i} {(x_{n} - a)}^{i - k - 2} \\ \Rightarrow & | a_{k + 1} | & = | (x_{n} - a) \sum_{i = k + 2}^{\infty} a_{i} {(x_{n} - a)}^{i - k - 2} | \\ \leq | x_{n} - a | \cdot \frac{c}{s^{k + 2}} \end{array}$
The equality $a_{k + 1} = 0$ is again due to the convergence $x_{n} \to a$ .

Consider:

[5.11.14] is often used for the so called comparison of coefficients:

If we have two convergent power series $(\sum_{i = 0}^{n} a_{i} {(X - a)}^{i})$ and $(\sum_{i = 0}^{n} b_{i} {(X - a)}^{i})$ and if we can find a sequence $(x_{n})$ in the intersection of their domains of convergence such that $a \neq x_{n} \to a$ and $\sum_{i = 0}^{\infty} a_{i} {(x_{n} - a)}^{i} = \sum_{i = 0}^{\infty} b_{i} {(x_{n} - a)}^{i}$ for all n, then both power series already coincide, i.e.

a_{k} = b_{k} for all k \in ℕ

[5.11.15]

Proof: The power series $(\sum_{i = 0}^{n} (a_{i} - b_{i}) {(X - a)}^{i})$ (cf. [5.11.17]) satisfies the condition in [5.11.14].

Next we show that the convergence of power series is compatible with basic arithmetics.

Proposition: Let $(\sum_{i = 0}^{n} a_{i} {(X - a)}^{i})$ and $(\sum_{i = 0}^{n} b_{i} {(X - a)}^{i})$ be two convergent power series and let $r_{1} and r_{2}$ be their radii of convergence respectively. Then their sum, their difference and their product are convergent power series as well and for $| x - a | < \min {r_{1}, r_{2}}$ we additionally have:

$\sum_{i = 0}^{\infty} (a_{i} + b_{i}) {(x - a)}^{i} = \sum_{i = 0}^{\infty} a_{i} {(x - a)}^{i} + \sum_{i = 0}^{\infty} b_{i} {(x - a)}^{i}$	[5.11.16]
$\sum_{i = 0}^{\infty} (a_{i} - b_{i}) {(x - a)}^{i} = \sum_{i = 0}^{\infty} a_{i} {(x - a)}^{i} - \sum_{i = 0}^{\infty} b_{i} {(x - a)}^{i}$	[5.11.17]
$\sum_{i = 0}^{\infty} (\sum_{j = 0}^{i} a_{j} b_{i - j}) {(x - a)}^{i} = \sum_{i = 0}^{\infty} a_{i} {(x - a)}^{i} \cdot \sum_{i = 0}^{\infty} b_{i} {(x - a)}^{i}$	[5.11.18]
If $b_{0} \neq 0$ the quotient $\frac{(\sum_{i = 0}^{n} a_{i} {(X - a)}^{i})}{(\sum_{i = 0}^{n} b_{i} {(X - a)}^{i})}$ is a convergent power series.	[5.11.19]

Proof: As series are sequences as well the first three assertions are only variants of the frist three limit theorems [5.6.1] - [5.6.3].

For 4. it is sufficient to represent only the quotient $\frac{1}{(\sum_{i = 0}^{n} b_{i} {(X - a)}^{i})}$ as a convergent power series. We begin with the construction of a power series $(\sum_{i = 0}^{n} c_{i} {(X - a)}^{i})$ such that

(\sum_{i = 0}^{n} b_{i} {(X - a)}^{i}) \cdot (\sum_{i = 0}^{n} c_{i} {(X - a)}^{i}) = {(1)}_{n \geq 0}

[2]

To that end we need coefficients $c_{i}$ which satisfy $\sum_{j = 0}^{i} b_{j} c_{i - j} = {\begin{array}{l} 1, if i = 0 \\ 0, if i > 0 \end{array}$ .
We define by recursion

c_{0} ≔ \frac{1}{b_{0}} \land c_{i + 1} ≔ - \frac{1}{b_{0}} \sum_{j = 1}^{i + 1} b_{j} c_{i + 1 - j}

and indeed get [2], because:

$\sum_{j = 0}^{0} b_{j} c_{0 - j} = b_{0} c_{0} = 1$
$\sum_{j = 0}^{i + 1} b_{j} c_{i + 1 - j} = b_{0} c_{i + 1} + \sum_{j = 1}^{i + 1} b_{j} c_{i + 1 - j} = 0$

As $b_{0} \neq 0$ evaluating the limit function $\sum_{i = 0}^{\infty} b_{i} {(X - a)}^{i}$ at the expansion point a will result in a value $\neq 0$ . We will prove now that this remains true in a complete neighbourhood $] a - \frac{1}{n}, a + \frac{1}{n} [$ of a: Suppose for every natural number $n > \frac{1}{r_{2}}$ there is an $x_{n}$ with $| x_{n} - a | < \frac{1}{n}$ such that $\sum_{i = 0}^{\infty} b_{i} {(x_{n} - a)}^{i} = 0$ . But as $a \neq x_{n} \to a$ [5.11.14] provides us with $b_{k} = 0$ for all k, a clearly contadicting $b_{0} \neq 0$ .

So there is a neighbourhood of a with no zeros for the limit function $\sum_{i = 0}^{\infty} b_{i} {(X - a)}^{i}$ . For each x in this neighbourhood the forth limit theorem is thus applicable in [2]:

\sum_{i = 0}^{n} c_{i} {(x - a)}^{i} = \frac{1}{\sum_{i = 0}^{n} b_{i} {(x - a)}^{i}} \to \frac{1}{\sum_{i = 0}^{\infty} b_{i} {(x - a)}^{i}}

So $(\sum_{i = 0}^{n} c_{i} {(X - a)}^{i})$ converges and due to [2] represents the reciprocal $\frac{1}{(\sum_{i = 0}^{n} b_{i} {(X - a)}^{i})}$ .

By definition power series come with a prefixed expansion point. Surprisingly this expansion point could be replaced (locally) with any other point from the domain of convergence. The following theorem plays a technical key role for some of our future work.

Proposition (Rearrangement Theorem): Let $(\sum_{i = 0}^{n} a_{i} {(X - a)}^{i})$ be any convergent power series and $r > 0$ its radius of convergence. Then for every $b \in] a - r, a + r [$ the following assertion holds:

There is a convergent power series $(\sum_{i = 0}^{n} b_{i} {(X - b)}^{i})$ such that

\sum_{i = 0}^{\infty} b_{i} {(x - b)}^{i} = \sum_{i = 0}^{\infty} a_{i} {(x - a)}^{i} for all x with | x - b | < s ≔ r - | b - a |

[5.11.20]

Proof: Please note firstly that $s > 0$ . Furthermore we have for all $| x - b | < s$ :

| x - a | \leq | b - a | + | x - b | < | b - a | + s = r

[3]

We use the generalized binomial theorem [5.2.5] to calculate for those x:

\begin{array}{l} \sum_{i = 0}^{n} a_{i} {(x - a)}^{i} \\ = & \sum_{i = 0}^{n} a_{i} {((b - a) + (x - b))}^{i} \\ = & \sum_{i = 0}^{n} \sum_{j = 0}^{i} a_{i} (\begin{matrix} i \\ j \end{matrix}) {(b - a)}^{i - j} {(x - b)}^{j} . \end{array}

[4]

So we need to study the double series $(\sum_{i = 0}^{n} \sum_{j = 0}^{m} b_{i j})$ , where $b_{i j} ≔ {\begin{array}{l} a_{i} (\begin{matrix} i \\ j \end{matrix}) {(b - a)}^{i - j} {(x - b)}^{j}, if i \geq j \\ 0, if i < j \end{array}$

Due to [3] and according to [5.11.10] the series $(\sum_{i = 0}^{n} | a_{i} | {(| b - a | + | x - b |)}^{i})$ converges. Hence the following estimate holds for all $n, m \in ℕ$ :

\begin{array}{l} \sum_{i = 0}^{n} \sum_{j = 0}^{m} | b_{i j} | & \leq \sum_{i = 0}^{n} \sum_{j = 0}^{i} | b_{i j} | \\ = \sum_{i = 0}^{n} | a_{i} | {(| b - a | + | x - b |)}^{i} \\ \leq \sum_{i = 0}^{\infty} | a_{i} | {(| b - a | + | x - b |)}^{i} \end{array}

This proves the boundedness (and consequently the convergence) of the double series $(\sum_{i = 0}^{n} \sum_{j = 0}^{m} | b_{i j} |)$ . When calculating the limit of the double series $(\sum_{i = 0}^{n} \sum_{j = 0}^{m} b_{i j})$ [5.10.26] thus allows us to interchange the order of row limit and column limit (consider also [3] and [4]):

\begin{array}{l} \sum_{i = 0}^{\infty} a_{i} {(x - a)}^{i} & = \sum_{i = 0}^{\infty} \sum_{j = 0}^{i} b_{i j} \\ = \sum_{i = 0}^{\infty} \sum_{j = 0}^{\infty} b_{i j} \\ = \sum_{j = 0}^{\infty} \sum_{i = 0}^{\infty} b_{i j} \\ = \sum_{j = 0}^{\infty} \underset{≕ b_{j}}{\underset{︸}{\sum_{i = j}^{\infty} a_{i} (\begin{matrix} i \\ j \end{matrix}) {(b - a)}^{i - j}}} {(x - b)}^{j} \end{array}

5.10. 5.12.