Infinite Series

Johann Christian Vollerdt (1708-1769) was a German painter.

We have sections

• Infinite Series of Real Numbers 
• Comparison Test and Integral Test
• Taylor Polynomials and Taylor Series
• Differentiation and Integration of Power Series

\begin{equation}{\label{a}}\tag{A}\mbox{}\end{equation}

Infinite Series of Real Numbers

Consider the infinite series \({\displaystyle \sum_{k=0}^{\infty} a_{k}}\). Let \({\displaystyle s_{n}=\sum_{k=0}^{n} a_{k}}\) be the partial sum of the infinite series. When the sequence of partial sums \(\{s_{n}\}_{n=1}^{\infty}\) is convergent to a finite limit \(L\), the series \({\displaystyle \sum_{k=0}^{\infty} a_{k}}\) is said to converge to \(L\), which is also written as \({\displaystyle \sum_{k=0}^{\infty} a_{k}=L}\).
The number \(L\) is called the sum of the series. When the sequence of partial sums is divergent, the series \({\displaystyle \sum_{k=0}^{\infty} a_{k}}\) is daid to be divergent.

Example. We have some interesting examples.

(i) Consider the series
\[\sum_{k=0}^{\infty}\frac{1}{(k+1)(k+2)}.\]
Since
\[\frac{1}{(k+1)(k+2)}=\frac{1}{k+1}-\frac{1}{k+2},\]
the partial sum is given by
\[s_{n}=1-\frac{1}{n+2},\]
which means
\[\sum_{k=0}^{\infty}\frac{1}{(k+1)(k+2)}=1.\]

(ii) The following series are divergent
\[\sum_{k=0}^{\infty} 2^{k}\mbox{ and }\sum_{k=0}^{\infty} (-1)^{k}.\]

The following series is called geometric series
\[\sum_{k=0}^{\infty} x^{k}.\]
Then, we have following fundamental results

• For \(|x|<1\),  we have \({\displaystyle \sum_{k=0}^{\infty} x^{k}=\frac{1} {1-x}}\);
• For  \(|x}\geq 1\), the series \({\displaystyle \sum_{k=0}^{\infty} x^{k}}\) is divergent,  since the partial sum
  \[s_{n}=\frac{1-x^{n+1}}{1-x}\] is divergent.

Furthermore, for \(a\neq 0\), we have
\[a+ar+ar^{2}+\cdots +ar^{n}+\cdots =\left\{\begin{array}{ll}
a/(1-r) & \mbox{if \(|r|<1\)}\\
\mbox{diverges} & \mbox{if \(|r|\geq 1\)}.
\end{array}\right .\]

Theorem. We have the following properties.

(i) Suppose that the series

\[\sum_{k=0}^{\infty} a_{k}=L\mbox{ and }\sum_{k=0}^{\infty}b_{k}=M\]

are convergent. Then. the series

\[\sum_{k=0}^{\infty} (a_{k}+b_{k})=L+M\]

is convergent.

(ii) Suppose that the series

\[\sum_{k=0}^{\infty} a_{k}=L\]

is convergent. Then, the series

\[\sum_{k=0}^{\infty} \alpha a_{k}=\alpha L\]

is also convergent for each \(\alpha\in\mathbb{R}\). \(\sharp\)

Theorem. Let \(j\) be a positive integer. Then, the series \(\sum_{k=0}^{\infty} a_{k}\) is convergent if and only if the series \(\sum_{k=j}^{\infty} a_{k}\) is convergent. Moreover, we have that

\[\sum_{k=0}^{\infty} a_{k}=L\mbox{ implies }\sum_{k=j}^{\infty} a_{k}=L-(a_{0}+a_{1}+\cdots +a_{j-1})\]

or

\[\sum_{k=j}^{\infty} a_{k}=M\mbox{ implies }\sum_{k=0}^{\infty} a_{k}=M+(a_{0}+a_{1}+\cdots +a_{j-1}).\]

\begin{equation}{\label{t1}}\tag{1}\mbox{}\end{equation}

Theorem \ref{t1}.  Suppose that the series \(\sum_{k=0}^{\infty} a_{k}\) is convergent. Then, we have \(a_{k}\rightarrow 0\) as \(k\rightarrow\infty\). Equivalently, when \(a_{k}\) does not converge to zero as \(k\rightarrow\infty\), the series \(\sum_{k=0}^{\infty} a_{k}\) is divergent.

\begin{equation}{\label{b}}\tag{B}\mbox{}\end{equation}

Comparison Test and Integral Test.

Theorem. (The Integral Test). Suppose that the function \(f\) is continuous, decreasing, and positive on \([1,\infty )\). Then
\[\sum_{k=1}^{\infty} f(k)\mbox{ is convergent if and only if the improper integral }\int_{1}^{\infty}f(x)dx\mbox{ is convergent.}\]

Example. We have some interesting examples.

(i) The harmonic series
\[\sum_{k=1}^{\infty} \frac{1}{k}=1+\frac{1}{2}+\frac{1}{3}+\cdots\]
is divergent, since the function \(f(x)=1/x\) is continuous, decreasing, and positive on \([1,\infty )\) and the improper integral \(\int_{1}^{\infty}dx/x\) is divergent.

(ii) The \(p\)-series
\[\sum_{k=1}^{\infty} \frac{1}{k^{p}}=1+\frac{1}{2^{p}}+\frac{1}{3^{p}}+\cdots\mbox{ is convergent if and only if }p>1,\]
since the improper integral
\[\int_{1}^{\infty}\frac{dx}{x^{p}}=\left [\frac{x^{1-p}}{1-p}\right ]_{1}^{\infty}\mbox{ is convergent if and only if }p>1.\]

Example. We are going to show that the following series
\[\sum_{k=1}^{\infty}\frac{1}{k\ln (k+1)}=\frac{1}{\ln 2}+\frac{1}{2\ln 3}+\frac{1}{3\ln 4}+\cdots\]
is divergent. We begin by setting
\[f(x)=\frac{1}{x\ln (x+1)}.\]
Since \(f\) is continuous, decreasing, and positive on \([1,\infty )\), we can use the integral test by calculating
\begin{align*} \int_{1}^{b}\frac{dx}{x\ln (x+1)} & >\int_{1}^{b}\frac{dx}{(x+1)\ln (x+1)}\\ & =\left [\ln (\ln (x+1))\right ]_{1}^{b}=\ln (\ln (b+1))-\ln\ln 2.\end{align*}
As \(b\rightarrow\infty\), we have \(\ln\ln (b+1)\rightarrow\infty\), which shows that the improper integral
\[\int_{1}^{\infty}\frac{dx}{x\ln (x+1)}\]
is divergent. It follows that the series is divergent.

Theorem. (The Basic Comparison Test). Let \(\sum a_{k}\) be a series with the nonnegative terms.

(i) The series \(\sum a_{k}\) is convergent if there exists a convergent series \(\sum c_{k}\) with nonnegative terms satisfying \(a_{k}\leq c_{k}\) for all  sufficiently large \(k\) .

(ii) The series \(\sum a_{k}\) is divergent if there exists a divergent series \(\sum d_{k}\) with nonnegative terms satisfying \(d_{k}\leq a_{k}\) for all sufficiently large \(k\). \(\sharp\)

Example. We have some interesting examples.

(i) Since the series \({\displaystyle\sum\frac{1}{k^{3}}}\) is convergent, the series \({\displaystyle \sum\frac{1}{2k^{3}+1}}\) is also convergent by comparison with \[\frac{1}{2k^{3}+1}<\frac{1}{k^{3}}\]

(ii) Since the series \[\frac{1}{3(k+1)}=\frac{1}{3}\sum\frac{1}{k+1}\] is divergent, the series \({\displaystyle \sum\frac{1}{3k+1}}\) is also divergent by comparison with \[\frac{1}{3(k+1)}<\frac{1}{3k+1}.\]

(iii) Since the series \({\displaystyle \sum\frac{1}{k^{2}}}\) is convergent, the series \({\displaystyle \sum\frac{k^{3}}{k^{5}+5k^{4}+7}}\) is also convergent by comparison with \[\frac{k^{3}}{k^{5}+5k^{4}+7}<\frac{k^{3}}{k^{5}}.\]

(iv) Show that \({\displaystyle \sum\frac{1}{\ln (k+6)}}\) is divergent. Since \(\ln k/k\rightarrow 0\) as \(k\rightarrow\infty\), it follows  \(\ln (k+6)/(k+6)\rightarrow 0\). Therefore , we obtain
\[\frac{\ln (k+6)}{k}=\frac{\ln (k+6)}{k+6}\cdot\frac{k+6}{k}\rightarrow 0.\]
For sufficiently large \(k\),  we have \(\ln (k+6)/k<1\), which implies
\[\ln (k+6)<k\mbox{ and }\frac{1}{k}<\frac{1}{\ln (k+6)}.\]
Since the series \({\displaystyle \sum\frac{1}{k}}\) is divergent, we conclude that the series \({\displaystyle \sum\frac{1}{\ln (k+6)}}\) is divergent. \(\sharp\)

Theorem. (The Limit Comparison Test). Consider the series \(\sum a_{k}\) and \(\sum b_{k}\) with positive terms. Suppose that \(a_{k}/b_{k}\rightarrow L\) as \(k\rightarrow\infty\), where \(L\) is some positive number ($L\neq 0$ or \(\infty\)). Then either both series are convergent or
both series are divergent. \(\sharp\)

Example. We have some interesting examples.

(i) Consider the series
\[\sum\frac{3k^{2}+2k+1}{k^{3}+1}.\]
For sufficiently large \(k\), \({\displaystyle \frac{3k^{2}+2k+1}{k^{3}+1}}\) differs little from \({\displaystyle \frac{3k^{2}}{k^{3}}=\frac{3}{k}}\). We have
\[\frac{3k^{2}+2^{k}+1}{k^{3}+1}\div\frac{3}{k}=\frac{3k^{3}+2k^{2}+k}{3k^{3}+3}=\frac{1+2/(3k)+1/(3k^{2})}{1+1/k^{3}}\rightarrow 1\mbox{ as }k\rightarrow\infty.\]
Since the series \({\displaystyle \sum\frac{3}{k}}\) is divergent, it follows that the series is  divergent.

(ii) Consider the series
\[\sum\frac{5\sqrt{k}+100}{2k^{2}\sqrt{k}+9\sqrt{k}}.\]
For sufficiently large \(k\), \({\displaystyle \frac{5\sqrt{k}+100}{2k^{2}\sqrt{k}+9\sqrt{k}}}\) differs little from \({\displaystyle \frac{5\sqrt{k}} {2k^{2}\sqrt{k}}=\frac{5}{2k^{2}}}\). We have
\[\frac{5\sqrt{k}+100}{2k^{2}\sqrt{k}+9\sqrt{k}}\div\frac{5}{2k^{2}}=
\frac{10k^{2}\sqrt{k}+200k^{2}}{10k^{2}\sqrt{k}+45\sqrt{k}}=\frac
{1+20/\sqrt{k}}{1+9/2k^{2}}\rightarrow 1\mbox{ as }k\rightarrow\infty.\]
Since the series \({\displaystyle \sum\frac{5}{2k^{2}}}\) is convergent, it follows that the series is convergent.

(iii) Consider the series
\[\sum \sin\frac{\pi}{k}.\]
Recall \({\displaystyle \frac{\sin x}{x}\rightarrow 1}\) as \(x\rightarrow 0\). Since \(\pi /k\rightarrow 0\) as \(k\rightarrow\infty\), it follows
\[\frac{\sin (\pi /k)}{\pi/k}\rightarrow 1\mbox{ as }k\rightarrow\infty.\]
Since the series \(\sum (\pi /k)\) is divergent, it follows that the series \(\sum\sin (\pi /k)\) is divergent. \(\sharp\)

Theorem. (The Root Test). Let \(\sum a_{k}\) be a series with nonnegative terms satisfying
\[\sqrt[k]{a_{k}}\rightarrow\rho\mbox{ as }k\rightarrow\infty .\]

(i) If \(\rho <1\), then \(\sum a_{k}\) converges.

(ii) If \(\rho >1\), then \(\sum a_{k}\) diverges.

(iii) If \(\rho =1\), then the test is inconclusive; the series may either converge or diverge. \(\sharp\)

Example. We have some interesting examples.

(i) Consider the series \({\displaystyle \sum\frac{1}{(\ln k)^{k}}}\). We have \(\sqrt[k]{a_{k}}=1/\ln k\rightarrow 0\) as \(k\rightarrow\infty\). Therefore, the series is convergent.

(ii) Consider the series \({\displaystyle \sum\frac{2^{k}}{k^{3}}}\). We have
\[\sqrt[k]{a_{k}}=2\left (\frac{1}{k}\right )^{3/k}=2\left (\frac{1}{k^{1/k}}\right )^{3}\rightarrow 2\cdot 1^{3}=2\mbox{ as }k\rightarrow\infty\]
since \(k^{\frac{1}{k}}\rightarrow 1\) as \(k\rightarrow\infty\). Therefore, the series is divergent.

(iii) Consider the series \({\displaystyle \sum\left (1-\frac{1}{k}\right )^{k}}\). We have \(\sqrt[k]{a_{k}}=1-1/k\rightarrow 1\) as \(k\rightarrow\infty\). Therefore, the root test is inconclusive. Since \(a_{k}=(1-1/k)^{k}\) converges to \(1/e\) not to \(0\), the series is divergent by Theorem \ref{t1} \(\sharp\)

Theorem. (The Ratio Test). Let \(\sum a_{k}\) be a series with positive terms satisfying
\[\frac{a_{k+1}}{a_{k}}\rightarrow\lambda\mbox{ as }k\rightarrow\infty .\]

(i) If \(\lambda <1\), then \(\sum a_{k}\) converges.

(ii) If \(\lambda >1\), then \(\sum a_{k}\) diverges.

(iii) If \(\lambda =1\), then the test is inconclusive; the series may either converge or diverge. \(\sharp\)

\begin{equation}{\label{ex30}}\tag{2}\mbox{}\end{equation}
Example \ref{ex30}. We have some interesting examples.

(i) The series \({\displaystyle \sum \frac{1}{k!}}\) is convergent, since
\[\frac{a_{k+1}}{a_{k}}=\frac{1}{(k+1)!}\cdot\frac{k!}{1}=\frac{1}{k+1}\rightarrow 0\mbox{ as }k\rightarrow\infty .\]

(ii) The series \({\displaystyle \sum \frac{k}{10^{k}}}\) is convergent, since
\[\frac{a_{k+1}}{a_{k}}=\frac{k+1}{10^{k+1}}\cdot\frac{10^{k}}{k}=\frac{1}{10}\frac{k+1}{k}\rightarrow\frac{1}{10}\mbox{ as }k\rightarrow\infty .\]

(iii) The series \({\displaystyle \sum\frac{k^{k}}{k!}}\) is divergent, since
\[\frac{a_{k+1}}{a_{k}}=\frac{(k+1)^{k+1}}{(k+1)!}\cdot\frac{k!}{k^{k}}+\left (1+\frac{1}{k}\right )^{k}\rightarrow e>1\mbox{ as }k\rightarrow\infty .\]

(iv) Consider the series \({\displaystyle \sum\frac{a}{2k+1}}\). We have
\[\frac{a_{k+1}}{a_{k}}=\frac{1}{2(k+1)+1}\cdot\frac{2k+1}{1}=\frac{2k+1}{2k+3}\rightarrow 1\mbox{ as }k\rightarrow\infty .\]
Therefore, the ratio test is inconclusive. However, comparison with harmonic series shows that the series is divergent,  since
\[\frac{1}{2k+1}\div\frac{1}{k}=\frac{k}{2k+1}\rightarrow\frac{1}{2}\mbox{ and }\sum\frac{1}{k}\mbox{ diverges}.\]

Theorem. If \(\sum |a_{k}|\) converges, then \(\sum a_{k}\) converges. \(\sharp\)

Definition. Two concepts of convergence are defined below.

• (Absolute Convergence). A series \(\sum a_{k}\) is absolutely convergent when the series \(\sum |a_{k}|\) is convergent.
• (Conditional Convergence). A series is conditionally convergent when it is convergent but \(\sum |a_{k}|\) is divergent. \(\sharp\)

For example, the series \({\displaystyle \sum_{k=1}^{\infty}\frac{(-1)^{k+1}} {k^{2}}}\) is absolutely convergent, since \(\sum 1/k^{2}\) is a convergent \(p\)-series.

Series in which the consecutive terms have opposite signs are called alternating series. For example, the series
\[1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\cdots =\sum_{k=0}^{\infty}\frac{(-1)^{k}}{k+1}.\]
and
\[-1+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{4}}-\cdots =\sum_{k=1}^{\infty} \frac{(-1)^{k}}{\sqrt{k}}\]
are alternating series. In general, an alternating series will either have the form
\[a_{0}-a_{1}+a_{2}-a_{3}+a_{4}-\cdots =\sum_{k=0}^{\infty} (-1)^{k}a_{k}\]
or the form
\[-a_{0}+a_{1}-a_{2}+a_{3}-a_{4}+\cdots =\sum_{k=0}^{\infty} (-1)^{k}a_{k},\]
where \(\{a_{k}\}_{k=1}^{\infty}\) is a sequence of positive numbers.

Theorem. (Alternating Series Test). Let \(\{a_{k}\}\) be a sequence of positive numbers. Suppose that the following conditions are satisfied.

• \(a_{k+1}<a_{k}\) for all \(k\); that is, if the sequence \(\{a_{k}\}\) is decreasing;
• \(a_{k}\rightarrow 0\) as \(k\rightarrow\infty\).

Then the alternating series \({\displaystyle \sum_{k=0}^{\infty} (-1)^{k}a_{k}}\) is convergent. \(\sharp\)

We can see that the serieses
\[\sum_{k=0}^{\infty}\frac{(-1)^{k}}{k+1}\mbox{ and }\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{\sqrt{k}}\]
are conditionally convergent, and that the serieses
\[\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k!}\] is absolutely convergent by part (i) of Example \ref{ex30} .

A rearrangement of a series \(\sum a_{k}\) is a series that exactly the same terms but in a different order. In 1867 Riemann published a theorem on rearrangement of series that present the differences between the absolute convergence and conditional convergence. According to this theorem all rearrangements of an absolutely convergent series converge absolutely to the same sum. In sharp contrast, a series that is only conditionally convergent can be rearranged to converge to any number we please. It can also be arranged to diverge to \(\infty\), or to diverge to \(-\infty\), or even to oscillate between any two bounds we choose.

Example. We have shown that the series
\[\sum_{k=0}^{\infty}\frac{(-1)^{k}}{k+1}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+\cdots\]
is conditionally convergent. We also have
\[1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+\cdots=\ln 2\]
Therefore, we can obtain
\[\frac{1}{2}-\frac{1}{4}+\frac{1}{6}-\frac{1}{8}+\frac{1}{10}-\frac{1}{12}+\cdots =\frac{1}{2}\ln 2.\]
by multiplying \(1/2\). Adding the above two series, we get a rearrangement of the given series with a new sum given by
\[1+\frac{1}{3}-\frac{1}{2}+\frac{1}{5}+\frac{1}{7}-\frac{1}{4}+\cdots=\frac{3}{2}\ln 2.\]

\begin{equation}{\label{c}}\tag{C}\mbox{}\end{equation}

Taylor Polynomials and Taylor Series.

Theorem. (Taylor’s Theorem). Suppose that \(f\) has \(n+1\) continuous derivatives on an open interval \(I\) containing the point \(a\). Then, for each \(x\in I\), we have
\[f(x)=f(a)+f'(a)(x-a)+\frac{f”(a)}{2!}(x-a)^{2}+\cdots +\frac{f^{(n)}(a)}{n!}(x-a)^{n}+R_{n+1}(x),\]
where
\[R_{n+1}(x)=\frac{1}{n!}\int_{a}^{x} f^{(n+1)}(t)(x-t)^{n}dt.\]

The polynomial
\[P_{n}(x)=f(a)+f'(a)(x-a)+\frac{f”(a)}{2!}(x-a)^{2}+\cdots +\frac{f^{(n)}(a)}{n!}(x-a)^{n}\]
is called the \(n\)th Taylor polynomial for \(f\) in powers of \((x-a)\).

Corollary. (Lagrange Formula for the Remainder). Suppose that \(f\) has \(n+1\) continuous derivatives on an open interval \(I\) containing \(a\). Let \(x\in I\) and let \(P_{n}\) be the \(n\)th Taylor polynomial for \(f\) in powers of \((x-a)\). Then, we have
\[R_{n+1}(x)=\frac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1},\]
where \(c\) is some number between \(a\) and \(x\). \(\sharp\)

Let \(x\in I\) with \(x\neq a\), and let \(J\) be the interval joining \(a\) and \(x\). Then, an estimate for the remainder can be written: as
\begin{equation}{\label{e8}}\tag{3}
R_{n+1}(x)\leq\left (\max_{t\in J} |f^{(n+1)}(t)|\right )\frac{|x-a|^{n+1}}{(n+1)!}.
\end{equation}
Suppose that \(f\) is infinitely differentiable on an open interval \(I\) containing \(a\). Then, we have
\[f(x)=\sum_{k=0}^{n}\frac{f^{(k)}(a)}{k!}(x-a)^{k}+R_{n+1}(x).\]
When \(R_{n+1}(x)\rightarrow 0\) as \(n\rightarrow\infty\), we have
\[f(x)=f(a)+f'(a)(x-a)+\frac{f”(a)}{2!}(x-a)^{2}+\cdots +\frac{f^{(n)}(a)}{n!}(x-a)^{n}+\cdots\]
or we write
\[f(x)=\sum_{k=0}^{\infty}\frac{f^{(k)}(a)}{k!}(x-a)^{k}.\]
This is known as the Taylor expansion of \(f(x)\) in powers of \(x-a\). The series on the right is called a Taylor series in \((x-a)\). For \(a=0\), the series is sometimes called Maclaurin series after Colin Maclaurin, a Scottish mathematician (1698-1746).

Example. For \(a=0\), the \(n\)th Taylor polynomial of the exponential function \(f(x)=e^{x}\) takes the form
\[P_{n}(x)=1+x+\frac{x^{2}}{2!}+\frac{x^{3}}{3!}+\cdots +\frac{x^{n}}{n!}.\]
We will show \(R_{n+1}(x)\rightarrow 0\) as \(n\rightarrow\infty\). Fix \(x\), let \(M\) be the maximum value of the exponential function on the interval \(J\) that joins \(0\) and \(x\). (If \(x>0\), then \(M=e^{x}\); if \(x<0\), then \(M=e^{0}=1\).) Since \(f^{(n+1)}(t)=e^{t}\) for all \(n\), we have \(\max_{t\in J}|f^{(n+1)}(t)|=M\) for all \(n\). Using (\ref{e8}), we have
\[|R_{n+1}(x)|\leq M\frac{|x|^{n+1}}{(n+1)!}.\]
Since \({\displaystyle \frac{|x|^{n+1}}{(n+1)!}\rightarrow 0}\) as \(n\rightarrow\infty\), it follows \(R_{n+1}(x)\rightarrow 0\) as \(n\rightarrow\infty\). Therefore, we can approximate \(e^{x}\) as closely as we wish by Maclaurin series, i.e.,
\[e^{x}=\sum_{k=0}^{\infty}\frac{x^{k}}{k!}=1+x+\frac{x^{2}}{2!}+\frac {x^{3}}{3!}+\cdots\] for all real \(x\). $\sharp$

Example. For \(a=0\), the Taylor polynomials of the sine function \(f(x)=\sin x\) are as follows
\begin{align*}
& P_{0}(x)=0\\
& P_{1}(x)=P_{2}(x)=x\\
& P_{3}(x)=P_{4}(x)=x-\frac{x^{3}}{3!}\\
& P_{5}(x)=P_{6}(x)=x-\frac{x^{3}}{3!}+\frac{x^{5}}{5!}\\
& P_{7}(x)=P_{8}(x)=x-\frac{x^{3}}{3!}+\frac{x^{5}}{5!}-\frac{x^{7}}{7!}
\end{align*}
For all \(n\) and all real \(t\), we have \(|f^{(n+1)}(t)|\leq 1\). From the remainder estimate,  it follows
\[|R_{n+1}(x)|\leq\frac{|x|^{n+1}}{(n+1)!}.\]
Therefore \(|R_{n+1}(x)|\rightarrow 0\) for all real \(x\) as \(n\rightarrow\infty\). The Maclaurin series can be used to approximate \(\sin x\) for any real number \(x\) as closely as we wish, i.e.,
\[\sin x=\sum_{k=0}^{\infty}\frac{(-1)^{k}}{(2k+1)!}x^{2k+1}=x-\frac{x^{3}}{3!}+\frac{x^{5}}{5!}-\frac{x^{7}}{7!}+\cdots.\]
for all real \(x\). Similarly, we have
\[\cos x=\sum_{k=0}^{\infty}\frac{(-1)^{k}}{(2k)!}x^{2k}=1-\frac{x^{2}}{2!}+\frac{x^{4}}{4!}-\frac{x^{6}}{6!}+\cdots\]

for all real \(x\). \(\sharp\)

Example. The function \(f(x)=\ln (1+x)\) is defined on \((-1,\infty )\) and has derivatives
\[f'(x)=\frac{1}{1+X}, \quad f^{\prime\prime}(x)=-\frac{1}{(1+x)^{2}}, \quad f^{\prime\prime\prime}(x)=\frac{2}{(1+x)^{3}}, \quad f^{(4)}(x)=-\frac{3!}{(1+x)^{4}},\mbox{ and so on}.\]
For \(k\geq 1\), we have
\[f^{(k)}(x)=(-1)^{k+1}\frac{(k-1)!}{(1+x)^{k}}.\]
For \(a=0\), the \(n\)th Taylor polynomial takes the form
\[P_{n}(x)=\sum_{k=1}^{n}(-1)^{k+1}\frac{x^{k}}{k}=x-\frac{x^{2}}{2}+\cdots +(-1)^{n+1}\frac{x^{n}}{n},\]
All we have to show is \(R_{n+1}(x)\rightarrow 0\) as \(n\rightarrow\infty\) for \(-1<x\leq 1\). Using Taylor’s theorem, we have
\[R_{n+1}(x)=\frac{1}{n!}\int_{0}^{x}f^{(n+1)}(t)(x-t)^{n}dx,\]
which also says
\[R_{n+1}(x)=\frac{1}{n!}\int_{0}^{x}(-1)^{n+2}\frac{n!}{(1+t)^{n+1}}(x-t)^{n}dt=(-1)^{n}\int_{0}^{x}\frac{(x-t)^{n}}{(1+t)^{n+1}}dt.\]
For \(0\leq x\leq 1\), we have
\[|R_{n+1}(x)|=\int_{0}^{x}\frac{(x-t)^{n}}{(1+t)^{n+1}}dt\leq\int_{0}^{x}(x-t)^{n}dt=\frac{x^{n+1}}{n+1}\rightarrow 0.\]
For \(-1<x<0\), we have
\[|R_{n+1}(x)|=\left |\int_{0}^{x}\frac{(x-t)^{n}}{(1+t)^{n+1}}dt\right |=\int_{x}^{0}\left (\frac{t-x}{1+t}\right )^{n}\frac{1}{1+t}dt.\]
Using the mean-value theorem for integrals, there exists a number \(x_{n}\) between \(x\) and \(0\) satisfying
\[\int_{x}^{0}\left (\frac{t-x}{1+t}\right )^{n}\frac{1}{1+t}dt=\left (\frac{x_{n}-x}{1+x_{n}}\right )^{n}\frac{1}{1+x_{n}}(-x).\]
Since \(-x=|x|\) and \(0<1+x<1+x_{n}\), we obtain
\[|R_{n+1}(x)|<\left (\frac{x_{n}+|x|}{1+x_{n}}\right )^{n}\left (\frac{|x|}{1+x}\right ).\]
Since \(|x|<1\) and \(x_{n}<0\), we have
\[x_{n}<|x|x_{n}\mbox{ and }x_{n}+|x|<|x|x_{n}+|x|=|x|(1+x_{n}),\]
which implies
\[\frac{x_{n}+|x|}{1+x_{n}}<|x|.\]
It follows
\[|R_{n+1}(x)|<|x|^{n}\left (\frac{|x|}{1+x}\right ).\]
Since \(|x|<1\) and \(R_{n+1}(x)\rightarrow 0\) as \(n\rightarrow\infty\),  we have
\[\ln (1+x)=\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k}x^{k}=x-\frac{x^{2}}{2}+\frac{x^{3}}{3}-\cdots\]

for \(-1<x\leq 1\). \(\sharp\)

Example.  Since
\[\ln (a+t)=\ln\left [a\left (1+\frac{t}{a}\right )\right ]=\ln a+\ln\left (1+\frac{t}{a}\right ),\]
From the previous example, we have
\[\ln\left (1+\frac{t}{a}\right )=\frac{t}{a}-\frac{1}{2}\left (\frac{t}{a}\right )^{2}+\frac{1}{3}\left (\frac{t}{a}\right )^{3}-\cdots\]
for \(-1<\frac{t}{a}\leq 1\) or \(-a<t\leq a\). Adding \(\ln a\) to both sides, we obtain
\[\ln (a+t)=\ln a+\frac{t}{a}-\frac{t^{2}}{2a^{2}}+\frac{t^{3}}{3a^{3}}-\cdots\] for \(-a<t\leq a\). Setting \(t=x-a\), we also obtain
\[\ln x=\ln a+\frac{x-a}{a}-\frac{(x-a)^{2}}{2a^{2}}+\frac{(x-a)^{3}}{3a^{3}}-\cdots\]
for all \(x\) satisfying \(-a<x-a\leq a\). Therefore, for \(a>0\) and \(0<x\leq 2a\), we finally obtain
\begin{align*}\ln x & =\ln a+\frac{x-a}{a}-\frac{(x-a)^{2}}{2a^{2}}+\frac{(x-a)^{3}}{3a^{3}}-\cdots\\ & =\ln a+\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{ka^{k}}(x-a)^{k}.\end{align*}

Example. Estimate \(e\) within \(0.001\). We have
\[|R_{n+1}(1)|\leq e\frac{|1|^{n+1}}{(n+1)!}<\frac{3}{(n+1)!}\]

where \(=e<3\). Since
\[\frac{3}{7!}=\frac{3}{5040}=\frac{1}{1680}<0.001,\]
we can take \(n=6\) to obtain
\[P_{6}(1)=1+1+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+\frac{1}{5!}+\frac{1}{6!}=\frac{1957}{720}\approx 2.7180556.\]

Example. Estimate \(\sin (0.5)\) within \(0.001\). We have
\[|R_{n+1}(0.5)|\leq\frac{(0.5)^{n+1}}{(n+1)!}.\]
Since
\[\frac{(0.5)^{5}}{5!}=\frac{1}{3840}<0.001,\]
we obtain
\[P_{4}(0.5)=P_{3}(0.5)=0.5-\frac{(0.5)^{3}}{3!}=\frac{23}{48}\approx0.4791666.\]

\begin{equation}{\label{d}}\tag{D}\mbox{}\end{equation}

Differentiation and Integration of Power Series.

The series of the form
\[\sum_{k=0}^{\infty} a_{k}x^{k}\mbox{ and }\sum_{k=0}^{\infty}a_{k}(x-a)^{k}\]
are called power series. Since a simple translation converts \(\sum a_{k}(x-a)^{k}\) into \(\sum a_{k}x^{k}\), we can focus the attention on power series of the form \(\sum a_{k}x^{k}\).

Definition. A power series \(\sum a_{k}x^{k}\) is said to converge at \(c\) when \(\sum a_{k}c^{k}\) is convergent. We say that the power series \(\sum a_{k}x^{k}\) is convergent on the set \(S\) when \(\sum a_{k}x^{k}\) is convergent for each \(x\in S\). \(\sharp\)

Theorem. If \(\sum a_{k}x^{k}\) converges at \(c\neq 0\), then it converges abosoultely for \(|x|<|c|\). If \(\sum a_{k}x^{k}\) diverges at \(c\), then it diverges for \(|x|>|c|\). \(\sharp\)

Now we consider the notion of radius of convergence.

• The series is convergent only at \(x=0\). This is what happens with \(\sum k^{k}x^{k}\). For \(x\neq 0\), since \({\displaystyle \lim_{k\rightarrow\infty} k^{k}x^{k}\neq 0}\), it says that the series cannot be convergent. In this case, we say that the radius of convergence is \(0\).
• The series is absolutely convergent for all real numbers \(x\). This is what happens with the exponential series \(\sum x^{k}/k!\). In this case, we say that the radius of convergence is \(\infty\).
• There exists a positive number \(r\) such that the series converges for \(|x|<r\) and diverges for \(|x|>r\). This is what happens with the geometric series \(\sum x^{k}\). In this case, we say that the radius of convergence is \(r\).

In general, the behavior of a power series at \(-r\) and \(r\) is not predictable. For example, the series
\[\sum x^{k},\sum\frac{(-1)^{k}}{k}x^{k},\sum\frac{1}{k}x^{k},\sum\frac{1}{k^{2}}x^{k}\]
all have radius of convergence \(1\), but the first series converges only on \((-1,1)\), the second series converges on \((-1,1]\), the third on \([-1,1)\), and the fourth on \([-1,1]\). The maximal interval on which a power series converges is called interval of convergence. For a series with infinite radius of convergence, the interval of convergence is \((-\infty ,\infty )\). For a series with radius of convergence \(r\), the interval of convergence can be \([-r,r]\), \((-r,r]\), \([-r,r)\) or \((-r,r)\). For a series with radius of convergence \(0\), the interval of convergence reduces to a point \(\{0\}\).

Example. Verify that the series \[\sum\frac{(-1)^{k}}{k}x^{k}\] has interval of convergence \((-1,1]\). We first consider the series
\[\sum\left |\frac{(-1)^{k}}{k}x^{k}\right |=\sum\frac{1}{k}|x|^{k}.\] Set \(b_{k}=|x|^{k}/k\) and note
\[\frac{b_{k+1}}{b_{k}}=\frac{k}{k+1}\frac{|x|^{k+1}}{|x|^{k}}=\frac{k}{k+1}|x|\rightarrow |x|\]  as \(k\rightarrow\infty\).
By the ratio test, the above series converges for \(|x|<1\) and diverges for \(|x|>1\). It follows that the original series converges absolutely for \(|x|<1\) and diverges for \(|x|>1\). The radius of convergence is therefore \(1\). Now, we test the endpoints \(x=-1\) and \(x=1\). At \(x=-1\), we see that
\[\sum\frac{(-1)^{k}}{k}(-1)^{k}=\sum\frac{1}{k}\] is a divergent harmonic series. At \(x=1\), we see that \({\displaystyle \sum\frac{(-1)^{k}}{k}}\) is a convergent alternating series. \(\sharp\)

Example. Verify that the series \({\displaystyle \frac{1}{k^{2}}x^{k}}\) has interval of convergence \([-1,1]\). We first consider the series
\[\sum\left |\frac{1}{k^{2}}x^{k}\right |=\sum\frac{1}{k^{2}}|x|^{k}.\] Set \(b_{k}=|x|^{k}/k^{2}\) and note
\[\frac{b_{k+1}}{b_{k}}=\frac{k^{2}}{(k+1)^{2}}\frac{|x|^{k+1}}{|x|^{k}}=\left (\frac{k}{k+1}\right )^{2}|x|\rightarrow |x|\] as \(k\rightarrow\infty\).
By the ratio test, the above series converges for \(|x|<1\) and diverges for \(|x|>1\). It follows that the original series converges absolutely for \(|x|<1\) and diverges for \(|x|>1\). The radius of convergence is therefore \(1\). Now, we test the endpoints \(x=-1\) and \(x=1\). At \(x=-1\), we see that \({\displaystyle \sum\frac{(-1)^{k}}{k^{2}}}\) is a convergent alternating series. At \(x=1\), we see that \({\displaystyle \sum\frac{1}{k^{2}}}\) is a convergent \(p\)-series. \(\sharp\)

Example.  Find the interval of convergence of \({\displaystyle \sum\frac{k}{6^{k}}x^{k}}\). We begin by examining the series \[\sum\left |
\frac{k}{6^{k}}x^{k}\right |=\sum\frac{k}{6^{k}}|x|^{k}.\]
Set \(b_{k}=k|x|^{k}/6^{k}\) and apply the root test (The ratio test will also work). Since
\[\sqrt[k]{b_{k}}=\frac{1}{6}k^{1/k}|x|\rightarrow\frac{1}{6}|x|\]
as \(k\rightarrow\infty\) (recall \(k^{1/k}\rightarrow 1\)), we can see that the above series converges for \(\frac{1}{6}|x|<1\), i.e., \(|x|<6\), and diverges for \(\frac{1}{6}|x|>1\), i.e., \(|x|>6\). Testing the endpoints, at \(x=6\), the series \({\displaystyle \sum\frac{k}{6^{k}}6^{k}=\sum k}\) is divergent, and at \(x=-6\), the series \({\displaystyle \sum\frac{k}{6^{k}}(-6)^{k}=\sum (-1)^{k}k}\) is divergent. Therefore, the interval of convergence is \((-6,6)\). \(\sharp\)

Example. Find the interval of convergence of \({\displaystyle \sum\frac{(2k)!}{(3k)!}x^{k}}\). We begin by examining the series
\[\sum\left |\frac{(2k)!}{(3k)!}x^{k}\right |=\sum\frac{(2k)!}{(3k)!}|x|^{k}.\]
Set \({\displaystyle b_{k}=\frac{(2k)!}{(3k)!}|x|^{k}}\). Since factorials are involved, we will use the ratio test. We have
\[\frac{b_{k+1}}{b_{k}}=\frac{[2(k+1)]!}{[3(k+1)]!}\frac{(3k)!}{(2k)!}\frac{|x|^{k+1}}{|x|^{k}}=\frac{(2k+2)(2k+1)}{(3k+3)(3k+2)(3k+1)}|x|.\]
Since
\[\frac{(2k+2)(2k+1)}{(3k+3)(3k+2)(3k+1)}\rightarrow 0\]  as \(k\rightarrow\infty\), the ratio \(b_{k+1}/b_{k}\) tends to zero no matter what \(x\) is. By the ratio test, the above series converges for all \(x\). Therefore, the original series converges absolutely for all \(x\), which also says that the radius of convergence is \(\infty\) and the interval of convergence is \((-\infty ,\infty )\). \(\sharp\)

Example. Find the interval of convergence of \({\displaystyle \sum (\frac{1}{2}k)^{k}x^{k}}\). Since \((\frac{1}{2}k)^{k}x^{k}\rightarrow 0\) only if \(x=0\), there is no need to invoke the ratio test or the root test. The series can converge only at \(x=0\). \(\sharp\)

Example. Find the interval of convergence of \({\displaystyle \sum\frac{(-1)^{k}}{k^{2}3^{k}}(x+2)^{k}}\). We consider the series
\[\sum\left |\frac{(-1)^{k}}{k^{2}3^{k}}(x+2)^{k}\right |=\sum\frac{|x+2|^{k}}{k^{2}3^{k}}.\]
Set \(b_{k}=|x+2|^{k}/k^{2}3^{k}\) and apply the ratio test. We obtain
\[\frac{b_{k+1}}{b_{k}}=\frac{k^{2}}{3(k+1)^{2}}|x+2|\rightarrow\frac{1}{3}|x+2|\] as \(k\rightarrow\infty\). Therefore, the series is absolutely convergent for \(\frac{1}{3}|x+2|<1\) or \(|x+2|<3\), which is the same as \(-5<x<1\). We now check the endpoints. At \(x=-5\), we see that \({\displaystyle \sum\frac{(-1)^{k}}{k^{2}3^{k}}(-3)^{k}=\sum\frac{1}{k^{2}}}\) is a convergent \(p\)-series. At \(x=1\), we see that \({\displaystyle \sum\frac{(-1)^{k}}{k^{2}3^{k}}3^{k}=\sum\frac{(-1)^{k}}{k^{2}}}\) is a convergent alternating series. Therefore, the interval of convergence is \([-5,1]\). \(\sharp\)

Theorem. Suppose that the series
\[\sum_{k=0}^{\infty} a_{k}x^{k}=a_{0}+a_{1}x+a_{2}x^{2}+\cdots +a_{n}x^{n}+\cdots\]
is convergent on \((-c,c)\). Then, the series
\[\sum_{k=0}^{\infty}\frac{d}{dx}(a_{k}x^{k})=\sum_{k=0}^{\infty} ka_{k}x^{k-1}=a_{1}+2a_{2}x+3a_{3}x^{2}+\cdots +na_{n}x^{n-1}+\cdots\]
is also convergent on \((-c,c)\). \(\sharp\)

Repeated application of the above theorem, we have
\[\sum_{k=0}^{\infty}\frac{d^{2}}{dx^{2}}(a_{k}x^{k}),\quad\sum_{k=0}^{\infty}\frac{d^{3}}{dx^{3}}(a_{k}x^{k}),
\quad\sum_{k=0}^{\infty}\frac{d^{4}}{dx^{4}}(a_{k}x^{k}),\mbox{ and so on,}\]
which are all convergent on \((-c,c)\).

Corollary. Suppose that the series \({\displaystyle \sum_{k=0}^{\infty} a_{k}x^{k}}\) has a radius of convergence \(r\). Then, each of the series
\[\sum_{k=0}^{\infty}\frac{d}{dx}(a_{k}x^{k}),\quad
\sum_{k=0}^{\infty}\frac{d^{2}}{dx^{2}}(a_{k}x^{k}),\quad
\sum_{k=0}^{\infty}\frac{d^{3}}{dx^{3}}(a_{k}x^{k}),\mbox{ and so on,}\]
has radius of convergence \(r\). \(\sharp\)

Even though \(\sum a_{k}x^{k}\) and its “derivative” \(\sum ka_{k}x^{k-1}\) have the same radius of convergence, their intervals of convergence may be different. For example, the interval of convergence of the series \({\displaystyle \sum_{k=1}^{\infty}\frac{1}{k^{2}}x^{k}}\) is \([-1,1]\), whereas the interval of convergence of its derivative \({\displaystyle \sum_{k=1}^{\infty}\frac{1}{k}x^{k-1}}\) is \([-1,1)\). Endpoints must be checked separately.

Theorem. (The Differentiability Theorem) Given
\[f(x)=\sum_{k=0}^{\infty} a_{k}x^{k}\mbox{ for all }x\in (-c,c),\]
the function \(f\) is differentiable on \((-c,c)\) given by
\[f'(x)=\sum_{k=0}^{\infty}\frac{d}{dx}(a_{k}x^{k})\mbox{ for all }x\in (-c,c).\]

By applying the above theorem to \(f’\), we can see that \(f’\) is itself differentiable. This in turn implies that \(f”\) is differentiable, and so on. In short, \(f\) has derivatives of all orders. This discussion up to this point can be summarized as follows. In the interior of its interval of convergence, a power series defines an infinitely differentiable function, the derivatives of which can be obtained by differentiating term by term given by
\[\frac{d^{n}}{dx^{n}}\left (\sum_{k=0}^{\infty} a_{k}x^{k}\right )=
\sum_{k=0}^{\infty}\frac{d^{n}}{dx^{n}}(a_{k}x^{k})\] for all \(n\).

Example. We can sum the series \({\displaystyle \sum_{k=1}^{\infty}\frac{x^{k}}{k}}\) for all \(x\in (-1,1)\) by setting \({\displaystyle g(x)=\sum_{k=1}^{\infty}\frac{x^{k}}{k}}\) for all \(x\in (-1,1)\). We also have
\[g'(x)=\sum_{k=1}^{\infty} x^{k-1}=\sum_{n=0}^{\infty} x^{n}=\frac{1}{1-x}.\]
With \(g'(x)=1/(1-x)\) and \(g(0)=0\), we obtain
\[g(x)=-\ln (1-x)=\ln\left (\frac{1}{1-x}\right ).\]
It follows
\[\sum_{k=1}^{\infty}\frac{x^{k}}{k}=\ln\left (\frac{1}{1-x}\right )\]
for all \(x\in (-1,1)\). \(\sharp\)

Theorem. (Term by Term Integration). Suppose that \({\displaystyle f(x)=\sum_{k=0}^{\infty}a_{k}x^{k}}\) converges on \((-c,c)\). Then \({\displaystyle g(x)=\sum_{0}^{\infty}\frac{a_{k}}{k+1}x^{k+1}}\) converges on \((-c,c)\) satisfying
\[\int f(x)dx=g(x)+C.\]

Term by term integration can be expressed as follows:
\[\int\left (\sum_{k=0}^{\infty} a_{k}x^{k}\right )dx=\sum_{k=0}^{\infty}\frac{a_{k}}{k+1}x^{k+1}+C.\]

From the above theorem, if \(\sum a_{k}x^{k}\) has radius of convergence \(r\), then its “integral” \(\sum [1/(k+1)]a_{k}x^{k+1}\) also has radius of convergence \(r\). As in the case of differentiating power series, convergence at the endpoints has to be tested separately. If a power series converges at \(c\) and converges at \(d\), then it converges at all number in between and
\[\int_{c}^{d}\left (\sum_{k=0}^{\infty} a_{k}x^{k}\right )dx=
\sum_{k=0}^{\infty}\left (\int_{c}^{d} a_{k}x^{k}dx\right )=
\sum_{k=0}^{\infty}\frac{a_{k}}{k+1}(d^{k+1}-c^{k+1}).\]

Theorem. (Abel’s Theorem). Suppose that the series \(f(x)=\sum a_{k}x^{k}\) converges on \((-c,c)\). If \(f\) is continuous at one of the endpoints ($c$
or \(-c\)) and the series converges there, then the series represents the function at that point. In other words, we have \(f(c)=\sum a_{k}c^{k}\) when \(f\) is continuous at \(c\) and the series \(\sum a_{k}c^{k}\) is convergent, and \(f(-c)=\sum a_{k}(-c)^{k}\) when \(f\) is continuous at \(-c\) and the series \(\sum a_{k}(-c)^{k}\) is convergent. \(\sharp\)

Example. Prove the expansion
\[\tan^{-1}x=x-\frac{x^{3}}{3}+\frac{x^{5}}{5}-\frac{x^{7}}{7}+\cdots\mbox{ for }-1\leq x\leq 1.\]
For \(x\in (-1,1)\), we have
\[\frac{1}{1+x^{2}}=\frac{1}{1-(-x^{2})}=\sum_{k=0}^{\infty} (-1)^{k}x^{2k}.\]
Taking integration, we obtain
\[\tan^{-1}x=\int\left (\sum_{k=0}^{\infty} (-1)^{k}x^{2k}\right )dx=\sum_{k=0}^{\infty}\frac{(-1)^{k}}{2k+1}x^{2k+1}+C.\]
The constant \(C=0\), since the series on the right and the inverse tangent are both zero at \(x=0\). Therefore, for all \(x\in (-1,1)\), we have
\[\tan^{-1}x=\sum_{k=0}^{\infty}\frac{(-1)^{k}}{2k+1}x^{2k+1}=x-\frac{x^{3}}{3}+\frac{x^{5}}{5}-\frac{x^{7}}{7}+\cdots.\]
This series also represents the function at \(x=-1\) and \(x=1\) following directly from the Abel’s theorem, since at both these points \(\tan^{-1}x\) is
continuous, and at both of these points the series converges. \(\sharp\)

Suppose that we are trying to evaluate \(\int_{a}^{b} f(x)dx\), but cannot find an antiderivative. If we can expand \(f(x)\) in a convergent power series, then we can estimate the integral by forming the series and integrating term by term.

Example. We will estimate
\[\int_{0}^{1} e^{-x^{2}}dx\]
by expanding the integral in a power series and integrating term by term. We know
\[e^{x}=1+x+\frac{x^{2}}{2!}+\frac{x^{3}}{3!}+\frac{x^{4}}{4!}+\cdots\mbox{ for all }x.\]
From this, we have
\[e^{-x^{2}}=1-x^{2}+\frac{x^{4}}{2!}-\frac{x^{6}}{3!}+\frac{x^{8}}{4!}-\cdots\mbox{ for all }x,\]
which implies
\begin{align*}
\int_{0}^{1} e^{-x^{2}} & =\left [x-\frac{x^{3}}{3}+\frac{x^{5}}{5(2!)}-\frac{x^{7}}{7(3!)}+\frac{x^{9}}{9(4!)}-\cdots\right ]_{0}^{1}\\
& =1-\frac{1}{3}+\frac{1}{5(2!)}-\frac{1}{7(3!)}+\frac{1}{9(4!)}-\cdots
\end{align*}
Similarly, we also have
\[\int_{0}^{1} e^{x^{2}}dx=1+\frac{1}{3}+\frac{1}{5(2!)}+\frac{1}{7(3!)}+\frac{1}{9(4!)}+\cdots\]

Next, we consider the binomial series. Consider the function \(f(x)=(1+x)^{\alpha}\), where \(\alpha\neq 0\) a real number. Let
\[\left (\begin{array}{c} \alpha\\ k\end{array}\right )=\frac{\alpha (\alpha -1)(\alpha -2)\cdots (\alpha -(k-1))}{k!}.\]
For example, if \(\alpha=3/2\) and \(k=3\), then
\[\left (\begin{array}{c} 3/2\\ 3\end{array}\right )=\frac{(3/2)(3/2-1)(3/2-2)}{3!}=-\frac{1}{16}.\]
We can show
\[(1+x)^{\alpha}=\sum_{k=0}^{\infty}\left (\begin{array}{c} \alpha\\ k\end{array}\right ) x^{k}=1+\alpha x+
\frac{\alpha (\alpha -1)}{2!}x^{2}+\frac{\alpha (\alpha -1)(\alpha -2)}{3!}x^{3}+\cdots\] for all \(x\in (-1,1)\). We can also show that the binomial series \({\displaystyle \sum_{k=0}^{\infty}\left (\begin{array}{c} \alpha\\ k\end{array}\right ) x^{k}}\) has radius of convergence \(1\) by using the ration test.