OBJECTIVES

When you finish this section, you should be able to:

1. Determine whether a power series converges (p. 600)
2. Find the interval of convergence of a power series (p. 603)
3. Define a function using a power series (p. 604)
4. Use properties of power series (p. 606)

spanDEFINITIONspan Power Series

If \(x\) is a variable, then a series of the form \[\bbox[5px, border:1px solid black, #F9F7ED]{\bbox[#FAF8ED,5pt]{ \sum\limits_{k\,=\,0}^{\infty}a_{k}x^{k}=a_{0}+\sum\limits_{k=1}^{\infty}a_{k}x^{k}=a_{0}+a_{1}x+a_{2}x^{2}+\cdots}} \]

where the coefficients \(a_{0}, a_{1}, a_{2},\ldots\) are constants, is called a power series in \({\bf{x}}\) or a power series centered at \({\bf 0}\).

A series of the form \[\bbox[5px, border:1px solid black, #F9F7ED]{\bbox[#FAF8ED,5pt]{ \sum\limits_{k\,=\,0}^{\infty}a_{k}(x-c)^{k}=a_{0}+\sum\limits_{k=1}^{\infty }a_{k}(x-c)^{k}=a_{0}+a_{1}(x-c)+a_{2}(x-c)^{2}+\cdots }} \]

where \(c\) is a constant, is called a power series in \({\bf {(x-c)}}\) or a power series centered at \({\bf c}\).

IN WORDS

A power series \(\sum\limits_{k\,=\,0}^{\infty}{a}_{k}{\,x}^{k}\) is a sum of an infinite number of monomials.

Determining Whether a Power Series Converges

Find all numbers \(x\) for which each power series in \(x\) converges.

1. \(\sum\limits_{k\,=\,0}^{\infty }\dfrac{x^{k}}{k!}=1+x+\dfrac{x^{2}}{2!}+\dfrac{x^{3}}{3!}+\cdots \)
2. \(\sum\limits_{k\,=\,0}^{\infty }\dfrac{kx^{k}}{4^{k}}=\dfrac{x}{4}+\dfrac{2x^{2}}{4^{2}}+\dfrac{3x^{3}}{4^{3}}+\cdots \)
3. \(\sum\limits_{k\,=\,0}^{\infty}k!x^{k}=1+x+2!x^{2}+3!x^{3}+\cdots \)

Solution (a) For the series \(\sum\limits_{k\,=\,0}^{\infty }\dfrac{x^{k}}{k!}\), we use the Ratio Test with \[ a_{n}=\frac{x^{n}}{n!}\qquad\hbox{and}\qquad a_{n+1}=\frac{x^{n+1}}{(n+1)!} \]

Then \[ \lim\limits_{n\rightarrow \infty }\left\vert \frac{a_{n+1}}{a_{n}}\right\vert =\lim\limits_{n\rightarrow \infty }\left\vert \frac{\dfrac{x^{n+1}}{(n+1) !}}{\dfrac{x^{n}}{n!}}\right\vert =\lim\limits_{n\rightarrow \infty } \frac{|x|^{n+1}\,n!}{(n+1)!\,|x|^{n}}=\vert x\vert \lim\limits_{n\rightarrow \infty }\frac{1}{n+1}=0 \]

Since the limit is less than \(1\) for every number \(x\), it follows from the Ratio Test that the power series \(\sum\limits_{k\,=\,0}^{\infty }\dfrac{x^{k}}{k!}\) is absolutely convergent for all real numbers.

(b) For \(\sum\limits_{k\,=\,0}^{\infty }\dfrac{kx^{k}}{4^{k}}\), we use the Ratio Test with \(a_{n}=\dfrac{nx^{n}}{4^{n}}\) and \(a_{n+1}=\dfrac{ (n+1)x^{n+1}}{4^{n+1}}\). Then \[ \begin{eqnarray*} \lim\limits_{n\,\rightarrow \,\infty }\left\vert \frac{a_{n+1}}{a_{n}} \right\vert &=& \lim\limits_{n\,\rightarrow \,\infty }\dfrac{\dfrac{ (n+1)\,\vert x\vert ^{n+1}}{4^{n+1}}}{\dfrac{n\,\,\vert x\vert ^{n}}{4^{n}}}=\lim\limits_{n\,\rightarrow \,\infty }\frac{ (n+1)\,|x|^{n+1}(4^{n})}{4^{n+1}\cdot n\,|x|^{n}} \\ &=& \vert x\vert \,\lim\limits_{n\,\rightarrow \,\infty }\frac{n+1}{4n}=\frac{|x|}{4} \end{eqnarray*} \]

By the Ratio Test, the series converges absolutely if \(\dfrac{|x|}{4} < 1\), or equivalently if \(\vert x\vert < 4\). It diverges if \(\dfrac{|x|}{4} \gt 1\) or equivalently if \(\vert x\vert >4\). The Ratio Test gives no information when \(\dfrac{|x|}{4}=1\), that is, when \(x=-4\) or \(x=4\). However, we can check these values directly by replacing \(x\) by \(4\) and \(-4\).

For \(x=4\), the series becomes \[ \sum_{k\,=\,0}^{\infty }\frac{k4^{k}}{4^{k}}=\sum_{k=1}^{\infty }k=1+2+\cdots \]

which diverges. For \(x=-4\), the series becomes \[ \sum_{k=0}^{\infty }\frac{k(-4)^{k}}{4^{k}}=\sum_{k=0}^{\infty }\frac{ (-1)^{k}k(4^{k})}{4^{k}}=\sum_{k=0}^{\infty }(-1)^{k}k=-1+2-3+\cdots \]

which also diverges. (Look at the sequence of partial sums.)

The series \(\sum\limits_{k\,=\,0}^{\infty }\dfrac{kx^{k}}{4^{k}}\) converges absolutely for \( -4< x< 4\) and diverges for \(\vert x\vert \geq 4\).

(c) For \(\sum\limits_{k\,=\,0}^{\infty }k!x^{k}\), we use the Ratio Test with \(a_{n}=n!\,x^{n}\) and \(a_{n+1}=(n+1)!\,x^{n+1}\). Then \[ \lim\limits_{n\,\rightarrow \,\infty }\left\vert \frac{a_{n+1}}{a_{n}} \right\vert =\lim\limits_{n\,\rightarrow \,\infty }\frac{(n+1)!\,\vert x\vert ^{n+1}}{n\dot{!}\,\vert x\vert ^{n}}=\vert x\vert\,\lim\limits_{n\,\rightarrow \,\infty } (n+1) =\left\{ \begin{array}{ll} 0 & \hbox{if} \, x=0 \\[4pt] \infty & \hbox{if} \, x\neq 0 \end{array} \right. \]

We conclude that the power series \(\sum\limits_{k\,=\,0}^{\infty }k!x^{k}\) converges only when \(x=0\). For any other number \(x\), the power series diverges.

NOW WORK

Problem 15.

THEOREM Convergence/Divergence of a Power Series

1. If the power series \(\sum\limits_{k\,=\,0}^{\infty }a_{k}x^{k}\) converges for a number \(x_{0}\neq 0\), then it converges absolutely for all numbers \(x\) for which \(\vert x\vert < \vert x_{0}\vert\).
2. If the power series \(\sum\limits_{k\,=\,0}^{\infty }a_{k}x^{k}\) diverges for a number \(x_{1}\), then it diverges for all numbers \(x\) for which \(\vert x\vert >\vert x_{1}\vert\).

RECALL

If a series \(\sum\limits_{k\,=\,0}^{\infty }{a}_{k}\) converges, then \(\lim\limits_{n\,\rightarrow \,\infty }a_{n}= 0\).

Proof

Part (a) Assume that \(\sum\limits_{k\,=\,0}^{\infty }a_{k}x_{0}^{k}\) converges. Then \[ \lim\limits_{n\rightarrow \infty }\left( a_{n}x_{0}^{n}\right) =0 \]

Using the definition of the limit of a sequence and choosing \(\varepsilon =1\) , there is a positive integer \(N\) for which \(\vert a_{n}x_{0}^{n} \vert < 1\) for all \(n > N\). Now for any number \(x\) for which \(\vert x\vert < |x_{0}|\), we have

The series \(\sum\limits_{k\,=\,0}^{\infty }\left\vert \dfrac{x}{x_{0}} \right\vert ^{k}\) is a convergent geometric series since \(r=\left\vert \dfrac{x}{x_{0}}\right\vert < 1 (\vert x\vert < \vert x_{0}\vert)\). Therefore, by the Comparison Test for Convergence, the series \(\sum\limits_{k\,=\,0}^{\infty }\left\vert a_{k}x^{k}\right\vert\) converges, and so the power series \(\sum\limits_{k\,=\,0}^{\infty }a_{k}x^{k}\) converges absolutely for all numbers \(x\) such that \(\vert x\vert ≤ \vert x_{0}\vert\).

Part (b) Suppose the series converges for some number \(x\), \( \vert x\vert >|x_{1}|\). Then it must converge for \(x_{1}\) [by Part (a)], which contradicts the hypothesis of the theorem. Therefore, the series diverges for all \(x\) such that \(\vert x\vert >|x_{1}|\).

THEOREM

For a power series \(\sum\limits_{k\,=\,0}^{\infty}a_{k} (x-c)^{k}\), exactly one of the following is true:

• The series converges only if \(x=c\).
• The series converges absolutely for all \(x\).
• There is a positive number \(R\) for which the series converges absolutely for all \(x\), \(\vert x-c\vert < R\), and diverges for all \(x,\) \(\vert x-c\vert >R\). The behavior of the series at \(\vert x-c\vert =R\) must be determined separately.

Finding the Interval of Convergence of a Power Series

Find the radius of convergence and the interval of convergence of the power series \[ \sum\limits_{k\,=\,1}^{\infty }\dfrac{x^{2k}}{k} \]

Solution We use the Ratio Test with \(a_{n}=\dfrac{x^{2n}}{n}\) and \(a_{n+1}=\dfrac{x^{2(n+1)}}{n+1}\). Then \[ \lim\limits_{n\,\rightarrow \,\infty}\left\vert \dfrac{a_{n+1}}{a_{n}}\right\vert =\lim\limits_{n\,\rightarrow \,\infty}\left\vert \dfrac{\dfrac{x^{2(n+1)}}{n+1}}{\dfrac{x^{2n}}{n}}\right\vert =\lim\limits_{n\,\rightarrow \,\infty}\left\vert \dfrac{\,x^{2n+2}}{n+1}\cdot \dfrac{n}{x^{2n}}\right\vert =x^{2}\lim\limits_{n\,\rightarrow \, \infty }\dfrac{n\,}{n+1}=x^{2} \]

The series converges absolutely if \(x^{2}< 1\), or equivalently, if \(-1< x< 1\).

To determine the behavior at the endpoints, we investigate \(x=-1\) and \(x=1\). When \(x=1\) or \(x=-1\), \(\dfrac{x^{2k}}{k}=\dfrac{( x^{2}) ^{k}}{k}= \dfrac{1^{k}}{k}=\dfrac{1}{k}\), so the series reduces to the harmonic series \(\sum\limits_{k\,=\,1}^{\infty }\dfrac{1}{k}\), which diverges. Consequently, the radius of convergence is \(R=1\), and the interval of convergence is \(-1< x< 1\), as shown in Figure 27.

NOW WORK

Problem 19(a).

Finding the Interval of Convergence of a Power Series

Find the radius of convergence \(R\) and the interval of convergence of the power series \[ \sum\limits_{k\,=\,0}^{\infty }\dfrac{x^{k}}{(k+2) ^{2k}} \]

Solution We use the Root Test. Then \[ \lim\limits_{n\rightarrow \infty }\sqrt[n]{\left|\dfrac{x^{n}}{(n+2) ^{2n}}\right|}=\lim\limits_{n \rightarrow \infty }\dfrac{\vert x\vert }{(n+2) ^{2}} =\vert x\vert \lim\limits_{n\rightarrow \infty }\dfrac{1}{(n+2) ^{2}}=0 \]

The series converges absolutely for all \(x\). The radius of convergence is \( R=\infty\), and the interval of convergence is \((-\infty ,\infty)\).

NOW WORK

Problems 19(b) and 21.

Finding the Interval of Convergence of a Power Series

Find the radius of convergence \(R\) and the interval of convergence of the power series \[ \sum\limits_{k\,=\,0}^{\infty}(-1)^{k}\dfrac{(x-2)^{k}}{k+1} \]

Solution \(\sum\limits_{k\,=\,0}^{\infty }(-1)^{k}\dfrac{(x-2)^{k}}{k+1 }\) is a power series centered at \(2\). We use the Ratio Test with \( a_{n}=(-1)^{n}\dfrac{(x-2)^{n}}{n+1}\) and \(a_{n+1}=(-1)^{n+1} \dfrac{(x-2)^{n+1}}{n+2}\). Then \[ \begin{eqnarray*} \lim\limits_{n\rightarrow \infty }\left\vert \dfrac{a_{n+1}}{a_{n}}\right\vert &=& \lim\limits_{n\rightarrow \infty }\left\vert \frac{\dfrac{(-1)^{n+1}(x-2) ^{n+1}}{n+2}}{\dfrac{(-1)^{n}\,(x-2) ^{n}}{n+1}} \right\vert =\lim\limits_{n\rightarrow \infty }\left\vert \frac{(n+1)\,(x-2) }{n+2}\right\vert \nonumber \\ &=& \vert x-2\vert \,\lim\limits_{n\rightarrow \infty }\dfrac{n+1}{n+2}=\vert x-2\vert \end{eqnarray*} \]

The series converges absolutely if \(\vert x-2\vert < 1\), or equivalently if \(1 < x < 3\). The radius of convergence is \(R=1\). We check the endpoints \(x=1\) and \(x=3\) separately.

If \(x=1\), \[ \begin{eqnarray*} \sum\limits_{k\,=\,0}^{\infty }(-1)^{k}\dfrac{(x-2)^{k}}{k+1}&=&\sum \limits_{k\,=\,0}^{\infty }(-1)^{k}\dfrac{(-1) ^{k}}{k+1} =\sum\limits_{k\,=\,0}^{\infty }\dfrac{(-1) ^{2k}}{k+1} =\sum\limits_{k\,=\,0}^{\infty }\dfrac{1}{k+1}\nonumber \\ &=&1+\frac{1}{2}+\frac{1}{3}+\cdots + \frac{1}{n+1}+\cdots \end{eqnarray*} \]

which is the divergent harmonic series.

If \(x=3\), \[ \sum\limits_{k\,=\,0}^{\infty }(-1)^{k}\dfrac{(x-2)^{k}}{k+1}=\sum \limits_{k\,=\,0}^{\infty }(-1)^{k}\dfrac{1}{k+1}=1-\frac{1}{2}+\frac{1}{3}- \frac{1}{4}+\cdots +\frac{(-1)^{n}}{n+1}+\cdots \]

which is the convergent alternating harmonic series.

The series \(\sum\limits_{k\,=\,0}^{\infty }(-1)^{k}\dfrac{(x-2) ^{k} }{k+1}\) converges for \(1 < x\leq 3\), as shown in Figure 28.

NOW WORK

Problem 31.

Analyzing a Function Defined by a Power Series

A function \(f\) is defined by the power series \(f(x)=\sum\limits_{k\,=\,0}^{\infty }x^{k}\).

1. Find the domain of \(f\).
2. Evaluate \(f\left(\dfrac{1}{2}\right)\) and \(f\left(-\dfrac{1}{3}\right)\).
3. Find \(f\).

Solution (a) \(\sum\limits_{k\,=\,0}^{\infty }x^{k}\) is a power series centered at \(0\) with \(a_{k}=1\). Then \[ f(x) =1+x+x^{2}+x^{3}+x^{4}+ \cdots \]

The domain of \(f\) equals the interval of convergence of the power series. Since the series \(\sum\limits_{k\,=\,0}^{\infty }x^{k}\) is a geometric series, it converges for \(\vert x\vert < 1\). The radius of convergence is \(1\), and the interval of convergence is \((-1,1)\). The domain of \(f\) is the open interval \((-1,1) \).

(b) The numbers \(\dfrac{1}{2}\) and \(-\dfrac{1}{3}\) are in the interval \((-1,1)\), so they are in the domain of \(f\). Then \(f\left( \dfrac{1}{2}\right)\) is a geometric series with \(r=\dfrac{1}{2}\), \(a=1\), and \[ f\left(\dfrac{1}{2}\right) =1+\dfrac{1}{2}+\left(\dfrac{1}{2}\right) ^{2}+\left(\dfrac{1}{2}\right) ^{3}+\cdots \,= \frac{a}{1-r} = \dfrac{1}{1-\dfrac{1}{2}}=2 \]

Similarly, \[ f\left( -\dfrac{1}{3}\right) =1-\dfrac{1}{3}+\left( -\dfrac{1}{3}\right) ^{2}+\left( -\dfrac{1}{3}\right) ^{3}+\cdots \,=\dfrac{1}{1+\dfrac{1}{3}}=\dfrac{3}{4} \]

(c) Since \(f\) is defined by a geometric series, we can find \(f\) by summing the series. \[ \begin{eqnarray*} &&f(x) =\sum\limits_{k\,=\,0}^{\infty }x^{k} = 1 + x + x^{2} + \cdots + x^{n} + \cdots \underset{\underset{\color{#00ADEE}{a=1; r = x }}{\color{#00ADEE}{\uparrow}}}{=}\dfrac{1}{1-x}\quad -1 < x < 1\\ \end{eqnarray*} \]

NOW WORK

Problem 45.

Representing a Function by a Power Series Centered at \(0\)

Represent each of the following functions by a power series centered at \(0\):

1. \(h(x) =\dfrac{1}{1-2x^{2}}\)
2. \(g(x) =\dfrac{1}{3+x}\)
3. \(F(x) =\dfrac{x^{2}}{1-x}\)

Solution We use the function \(f(x) = \dfrac{1}{1-x}\), \(-1< x < 1\), represented by the geometric series \( \sum\limits_{k\,=\,0}^{\infty }x^{k}\).

(a) In the geometric series for \(f(x) =\dfrac{1}{1-x} , \) we replace \(x\) by \(2x^{2}\). This series converges if \(\vert 2x^{2}\vert < 1\), or equivalently if \(-\dfrac{\sqrt{2}}{2} < x < \dfrac{\sqrt{2}}{2}\). Then on the open interval \(\left( -\dfrac{\sqrt{2}}{2},\,\dfrac{\sqrt{2}}{2}\right)\), the function \(h\,(x) =\dfrac{1}{ 1-2x^{2}}\) is represented by the power series \[ \begin{eqnarray*} h(x) &=& \dfrac{1}{1-2x^{2}}=1+(2x^{2}) + (2x^{2}) ^{2}+ (2x^{2})^{3}+ \cdots \\ &=& 1+2x^{2}+4x^{4}+8x^{6}+\cdots+2^{n}x^{2n} +\cdots=\sum\limits_{k\,=\,0}^{ \infty }(2x^{2}) ^{k}= \sum\limits_{k\,=\,0}^{\infty }2^{k}x^{2k} \end{eqnarray*} \]

(b) We begin by writing \[ g(x) =\dfrac{1}{3+x}=\dfrac{1}{3}\left(\dfrac{1}{1+\dfrac{x}{3}} \right) =\dfrac{1}{3}\left[ \dfrac{1}{1-\left( -\dfrac{x}{3}\right) }\right] \]

Now in the geometric series for \(f(x) =\dfrac{1}{1-x}\), replace \( x\) by \(-\dfrac{x}{3}\). This series converges if \(\left\vert -\dfrac{x}{3} \right\vert < 1\), or equivalently if \(-3< x < 3\). Then in the open interval \( (-3,3) \), \(g(x) =\dfrac{1}{3+x}\) is represented by the power series \[ \begin{eqnarray*} g\,(x) &=& \dfrac{1}{3+x}=\dfrac{1}{3}\left[ 1+\left( -\dfrac{x}{3} \right) +\left( -\dfrac{x}{3}\right) ^{2}+\left( -\dfrac{x}{3}\right) ^{3}+\cdots\right] =\dfrac{1}{3}\sum\limits_{k\,=0}^{\infty }(-1) ^{k}\left( \dfrac{x}{3}\right) ^{k}\nonumber \\ &=&\sum\limits_{k\,=0}^{\infty }\dfrac{ (-1) ^{k}x^{k}}{3^{k+1}} \end{eqnarray*} \]

(c) \(F(x) =\dfrac{x^{2}}{1-x}=x^{2}\left( \dfrac{1}{ 1-x}\right)\). Now for all numbers in the interval \((-1,\,1)\), \[ \dfrac{1}{1-x}=1+x+x^{2}+\cdots+x^{n}+\cdots \]

So for any number \(x\) in the interval of convergence, \(-1< x < 1\), we have \[ F(x) =x^{2}\left( 1+x+x^{2}+\cdots+x^{n}+\cdots\right) =x^{2}+x^{3}+x^{4}+\cdots+x^{n+2}+\cdots\,=\sum\limits_{k\,=2}^{\infty }x^{k} \]

NOTE

Notice that the power series for \(F\) begins at \(k=2\).

NOW WORK

Problem 55.

THEOREM Properties of Power Series

Let \(\sum\limits_{k\,=0}^{\infty }a_{k}x^{k}\) be a power series in \( x\) having a nonzero radius of convergence \(R\). Define the function \(f\) as \[ f(x)=\sum\limits_{k\,=\,0}^{\infty}a_{k}x^{k}=a_{0}+a_{1}x+a_{2}x^{2}+\cdots +a_{n}x^{n}+\cdots \quad -R< x< R \]

• Continuity property: \[\bbox[5px, border:1px solid black, #F9F7ED]{\bbox[#FAF8ED,5pt]{ \lim\limits_{x\,\rightarrow \,x_{0}}\left( \sum\limits_{k\,=\,0}^{\infty }a_{k}x^{k}\right) =\sum\limits_{k\,=\,0}^{\infty }\left( \lim\limits_{x\,\rightarrow \,x_{0}}a_{k}x^{k}\right) = \sum\limits_{k\,=\,0}^{\infty }a_{k}x_{0}^{k} \quad -R< x_{0}< R }} \]
• Differentiation property: \[\bbox[5px, border:1px solid black, #F9F7ED]{\bbox[#FAF8ED,5pt]{ \dfrac{d}{dx}\left( \sum\limits_{k\,=\,0}^{\infty }a_{k}x^{k}\right) =\sum\limits_{k\,=\,0}^{\infty }\left( \dfrac{d}{dx} a_{k}x^{k}\right) =\sum\limits_{k\,=\,1}^{\infty }k a_{k}x^{k-1} }} \]
• Integration property: \[\bbox[5px, border:1px solid black, #F9F7ED]{\bbox[#FAF8ED,5pt]{ \int_{0}^{x}\left(\sum\limits_{k\,=\,0}^{\infty }a_{k}t^{k}\right) dt=\sum\limits_{k\,=\,0}^{\infty }\left( \int_{0}^{x}a_{k}t^{k}~dt\right) =\sum\limits_{k\,=\,0}^{\infty }\dfrac{ a_{k}x^{k+1}}{k+1}}} \]

CAUTION

The theorem states that the radii of convergence of the three series \(\sum\limits_{k\,=\,0}^{\infty}a_{k}x^{k}\), \(\sum\limits_{k\,=\,1}^{\infty }ka_{k}x^{k-1}\), and \(\sum\limits_{k\,=\,0}^{\infty } \dfrac{a_{k}x^{k+1}}{k+1}\) are the same. This does not imply that the intervals of convergence are the same; the endpoints of the interval must be checked separately for each series. For example, the interval of convergence of \(\sum\limits_{k\,=\,0}^{\infty }\dfrac{x^{k}}{k}\) is \([-1,1)\), but the interval of convergence of its derivative \(\sum\limits_{k\,=\,1}^{\infty }x^{k-1}\) is \((-1,1)\).

Using the Differentiation Property of Power Series

Use the differentiation property of power series to find the derivative of \[ f(x) =\dfrac{1}{1-x}=\sum\limits_{k\,=\,0}^{\infty }x^{k} \]

Solution The function \(f(x) =\dfrac{1}{1-x}\), defined on the open interval \((-1,\,1) \), is represented by the power series \[ f(x) =\frac{1}{1-x}=1+x+x^{2}+ \cdots +x^{n}+\cdots = \sum\limits_{k\,=\,0}^{\infty }x^{k} \]

Using the differentiation property, we find that \[ f^{\prime} (x) =\frac{1}{(1-x)^{2}}=1+2x+3x^{2}+\cdots +nx^{n-1}+\cdots =\sum\limits_{k\,=1}^{\infty }kx^{k-1} \]

whose radius of convergence is \(1\).

NOW WORK

Problem 59(a).

Finding the Power Series Representation for \(\ln \dfrac{1}{1-x}\)

1. Find the power series representation for \(\ln \dfrac{1}{1-x}\).
2. Find \(\ln 2\).

Solution (a) If \(y=\ln \dfrac{1}{1-x}\), then \(y^\prime = \dfrac{1}{1-x}\). That is, \(y^\prime\) is represented by the geometric series \( \sum\limits_{k\,=\,0}^{\infty }x^{k}\), which converges on the interval \( (-1,\,1)\). So, if we use the integration property of power series for \(y^\prime =\dfrac{1}{1-x}\), we obtain a series for \(y=\ln \dfrac{1}{ 1-x}\). \[ \begin{eqnarray*} y^\prime &=&\dfrac{1}{1-x}=1+x+x^{2}+\cdots +x^{n}+\cdots = \sum\limits_{k\,=\,0}^{\infty }x^{k} \\ \int_{0}^{x}\dfrac{1}{1-t}dt &=&\int_{0}^{x}( 1+t+t^{2}+\cdots +t^{n}+\cdots) ~dt \nonumber\\ \ln \dfrac{1}{1-x} &=&x+\dfrac{x^{2}}{2}+\dfrac{x^{3}}{3}+\cdots+\dfrac{ x^{n+1}}{n+1}+\cdots\, =\sum\limits_{k\,=\,0}^{\infty }\dfrac{x^{k+1}}{k+1} \end{eqnarray*} \]

The radius of convergence of this series is \(1\).

To find the interval of convergence, we investigate the endpoints. When \( x=1, \) \[ x+\dfrac{x^{2}}{2}+\dfrac{x^{3}}{3}+\cdots+\dfrac{x^{n+1}}{n+1}+\cdots\,=1+ \dfrac{1}{2}+\dfrac{1}{3}+\cdots\, \]

the harmonic series, which diverges. When \(x=-1\), \[ x+\dfrac{x^{2}}{2}+\dfrac{x^{3}}{3}+\cdots+\dfrac{x^{n+1}}{n+1}+\cdots\,=-1+ \dfrac{1}{2}-\dfrac{1}{3}+\cdots+(-1) ^{n+1}\dfrac{1}{n+1}+\cdots \]

an alternating harmonic series, which converges. The interval of convergence of the power series \(\sum\limits_{k\,=\,1}^{\infty }\dfrac{x^{k}}{k}\) is \( [-1,\,1) \). So, \[ \begin{equation*} \bbox[5px, border:1px solid black, #F9F7ED]{\bbox[#FAF8ED,5pt]{\ln \dfrac{1}{1-x}=x+\dfrac{x^{2}}{2}+\dfrac{x^{3}}{3} +\cdots\,=\sum\limits_{k\,=\,0}^{\infty }\dfrac{x^{k+1}}{k+1}\qquad -1\leq x<1}} \tag{1} \end{equation*} \]

(b) To find \(\ln 2\), notice that when \(x=-1\), we have \(\ln \dfrac{1}{1-x} =\ln \dfrac{1}{2}=- \ln 2\). In (1) let \(x=-1\). Then \[ \begin{eqnarray*} -\ln 2 &=&-1+\dfrac{1}{2}-\dfrac{1}{3}+\cdots \nonumber \\ \ln 2 &=&1-\dfrac{1}{2}+\dfrac{1}{3}\,\cdots \end{eqnarray*} \]

NOW WORK

Problems 59(b) and 65.

Finding the Power Series Representation for \(\tan^{-1} {\bf x}\)

Show that the power series representation for \(\tan ^{-1}x\) is \[\bbox[5px, border:1px solid black, #F9F7ED]{\bbox[#FAF8ED,5pt]{ \tan ^{-1}x=x-\dfrac{x^{3}}{3}+\dfrac{x^{5}}{5}-\dfrac{ x^{7}}{7}+\cdots +(-1)^{n}\dfrac{x^{2n+1}}{2n+1}+\cdots =\sum\limits_{k\,=\,0}^{\infty }\dfrac{(-1) ^{k}x^{2k+1}}{2k+1} }} \]

Find the radius of convergence and the interval of convergence.

Solution If \(y=\tan ^{-1}x\), then \(y^\prime =\dfrac{1}{1+x^{2}}\), which is the sum of the geometric series \(\sum\limits_{k\,=\,0}^{\infty}(-1) ^{k}x^{2k}\). That is, \[ \dfrac{1}{1+x^{2}}=\sum\limits_{k\,=\,0}^{\infty }(-1) ^{k}x^{2k}=1-x^{2}+x^{4}-x^{6}+\cdots\, \]

This series converges when \(\vert x^{2}\vert < 1\), or equivalently for \(-1 < x < 1\). We use the integration property of power series to integrate \( y^\prime =\dfrac{1}{1+x^{2}}\) and obtain a series for \(y=\tan ^{-1}x\). \[ \begin{eqnarray*} \int_{0}^{x}\frac{dt}{1+t^{2}} &=&\int_{0}^{x}(1-t^{2}+t^{4}-\cdots)\, dt \nonumber \\ \tan ^{-1}x &=& x-\frac{x^{3}}{3}+\frac{x^{5}}{5}-\frac{x^{7}}{7}+\cdots +(-1)^{n}\frac{x^{2n+1}}{2n+1}+\cdots = \sum\limits_{k\,=\,0}^{\infty }(-1) ^{k}\dfrac{x^{2k+1}}{2k+1} \end{eqnarray*} \]

The radius of convergence is \(1\). To find the interval of convergence, we check \(x=-1\) and \(x=1\). For \(x=-1\), \[ -1+\frac{1}{3}-\frac{1}{5}+\frac{1}{7}-\cdots \]

For \(x=1\), we get \[ 1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\cdots \]

Both of these series satisfy the two conditions of the Alternating Series Test, and so each one converges. The interval of convergence is the closed interval \([-1,\,1] \).

NOW WORK

Problem 69.