1. One of the conventions we adopt is that \(x^0=1\) regardless of the value of \(x\text{.}\) Therefore \begin{equation*} \infser[0] x^n = 1+x+x^2+x^3+x^4+\ldots \end{equation*} This is a geometric series in \(x\) with \(r=x\text{.}\)

2. This series is centered at \(c=-1\text{.}\) Note how this series starts with \(n=1\text{.}\) We could rewrite this series starting at \(n=0\) with the understanding that \(a_0=0\text{,}\) and hence the first term is \(0\text{.}\) \begin{align*} \amp \infser (-1)^{n+1}\frac{(x+1)^n}n\\ \amp =(x+1) - \frac{(x+1)^2}{2} + \frac{(x+1)^3}{3} - \frac{(x+1)^4}{4}+\frac{(x+1)^5}{5}\ldots \end{align*}

3. This series is centered at \(c=\pi\text{.}\) Recall that \(0!=1\text{.}\) \begin{align*} \amp \infser[0] (-1)^{n+1} \frac{(x-\pi)^{2n}}{(2n)!} \\ \amp = -1+\frac{(x-\pi)^2}{2} - \frac{(x-\pi)^4}{24}+ \frac{(x-\pi)^6}{6!}-\frac{(x-\pi)^8}{8!}\ldots \end{align*}

1. We apply the Ratio Test to the series \(\ds \infser \abs{\frac{x^n}{n!}}\text{:}\) \begin{align*} \lim_{n\to\infty} \frac{\abs{x^{n+1}/(n+1)!}}{\abs{x^n/n!}} \amp = \lim_{n\to\infty} \abs{\frac{x^{n+1}}{x^n}\cdot\frac{n!}{(n+1)!}}\\ \amp = \lim_{n\to\infty} \abs{\frac x{n+1}}\\ \amp = 0 \text{ for all } x. \end{align*} The Ratio Test shows us that regardless of the choice of \(x\text{,}\) the series converges. Therefore the radius of convergence is \(R=\infty\text{,}\) and the interval of convergence is \((-\infty,\infty)\text{.}\)

2. We apply the Ratio Test to the series \(\ds \infser \abs{(-1)^{n+1}\frac{x^n}{n}} = \infser \abs{\frac{x^n}{n}}\text{:}\) \begin{align*} \lim_{n\to\infty} \frac{\abs{x^{n+1}/(n+1)}}{\abs{x^n/n}} \amp = \lim_{n\to\infty} \abs{\frac{x^{n+1}}{x^n}\cdot \frac{n}{n+1}}\\ \amp = \lim_{n\to\infty} n!x^n\\ \amp = \abs{x}. \end{align*} The Ratio Test states a series converges if the limit of \(\abs{a_{n+1}/a_n} = L\lt 1\text{.}\) We found the limit above to be \(\abs{x}\text{;}\) therefore, the power series converges when \(\abs{x} \lt 1\text{,}\) or when \(x\) is in \((-1,1)\text{.}\) Thus the radius of convergence is \(R=1\text{.}\) To determine the interval of convergence, we need to check the endpoints of \((-1,1)\text{.}\) When \(x=-1\text{,}\) we have the opposite of the Harmonic Series: \begin{align*} \infser (-1)^{n+1}\frac{(-1)^n}{n} \amp = \infser \frac{-1}{n}\\ \amp = -\infty. \end{align*} The series diverges when \(x=-1\text{.}\) When \(x=1\text{,}\) we have the series \(\ds \infser (-1)^{n+1}\frac{(1)^n}{n}\text{,}\) which is the Alternating Harmonic Series, which converges. Therefore the interval of convergence is \((-1,1]\text{.}\)

3. We apply the Ratio Test to the series \(\ds\infser[0] \abs{2^n(x-3)^n}\text{:}\) \begin{align*} \lim_{n\to\infty} \frac{\abs{ 2^{n+1}(x-3)^{n+1}}}{\abs{2^n(x-3)^n}} \amp = \lim_{n\to\infty} \abs{\frac{2^{n+1}}{2^n}\cdot\frac{(x-3)^{n+1}}{(x-3)^n}}\\ \amp =\lim_{n\to\infty} \abs{2(x-3)}. \end{align*} According to the Ratio Test, the series converges when \(\abs{2(x-3)}\lt 1 \implies \abs{x-3} \lt 1/2\text{.}\) The series is centered at 3, and \(x\) must be within \(1/2\) of 3 in order for the series to converge. Therefore the radius of convergence is \(R=1/2\text{,}\) and we know that the series converges absolutely for all \(x\) in \((3-1/2,3+1/2) = (2.5, 3.5)\text{.}\) We check for convergence at the endpoints to find the interval of convergence. When \(x=2.5\text{,}\) we have: \begin{align*} \infser[0] 2^n(2.5-3)^n \amp = \infser[0] 2^n(-1/2)^n\\ \amp =\infser[0] (-1)^n, \end{align*} which diverges. A similar process shows that the series also diverges at \(x=3.5\text{.}\) Therefore the interval of convergence is \((2.5, 3.5)\text{.}\)

4. We apply the Ratio Test to \(\ds \infser[0] \abs{n!x^n}\text{:}\) \begin{align*} \lim_{n\to\infty} \frac{\abs{ (n+1)!x^{n+1}}}{\abs{n!x^n}} \amp = \lim_{n\to\infty} \abs{(n+1)x}\\ \amp = \infty\ \text{ for all \(x\), except \(x=0\). } \end{align*} The Ratio Test shows that the series diverges for all \(x\) except \(x=0\text{.}\) Therefore the radius of convergence is \(R=0\text{.}\)

We find the derivative and indefinite integral of \(f(x)\text{,}\) following Theorem 8.6.7.

1. \begin{align*} \fp(x) \amp = \infser nx^{n-1} = 1+2x+3x^2+4x^3+\cdots\\ \amp = \infser[0](n+1)x^n\text{.} \end{align*} In Example 8.6.2, we recognized that \(\ds \infser[0] x^n\) is a geometric series in \(x\text{.}\) We know that such a geometric series converges when \(\abs{x}\lt 1\text{;}\) that is, the interval of convergence is \((-1,1)\text{.}\) To determine the interval of convergence of \(\fp(x)\text{,}\) we consider the endpoints of \((-1,1)\text{:}\) \begin{equation*} \fp(-1) = 1-2+3-4+\cdots, \text{ which diverges. } \end{equation*} \begin{equation*} \fp(1) = 1+2+3+4+\cdots, \text{ which diverges. } \end{equation*} Therefore, the interval of convergence of \(\fp(x)\) is \((-1,1)\text{.}\)

2. \(\ds F(x) = \int f(x)\ dx = C+\infser[0] \frac{x^{n+1}}{n+1} = C+ x+\frac{x^2}{2}+\frac{x^3}3+\cdots\) To find the interval of convergence of \(F(x)\text{,}\) we again consider the endpoints of \((-1,1)\text{:}\) \begin{equation*} F(-1) = C-1+1/2-1/3+1/4+\cdots \end{equation*} The value of \(C\) is irrelevant; notice that the rest of the series is an Alternating Series that whose terms converge to 0. By the Alternating Series Test, this series converges. (In fact, we can recognize that the terms of the series after \(C\) are the opposite of the Alternating Harmonic Series. We can thus say that \(F(-1) = C-\ln(2)\text{.}\)) \begin{equation*} F(1) = C+1+1/2+1/3+1/4+\cdots \end{equation*} Notice that this summation is \(C\ +\) the Harmonic Series, which diverges. Since \(F\) converges for \(x=-1\) and diverges for \(x=1\text{,}\) the interval of convergence of \(F(x)\) is \([-1,1)\text{.}\)

We start by making two notes: first, in Example 8.6.6, we found the interval of convergence of this power series is \((-\infty,\infty)\text{.}\) Second, we will find it useful later to have a few terms of the series written out: \begin{equation} \infser[0] \frac{x^n}{n!} = 1 + x + \frac{x^2}2+\frac{x^3}{6} + \frac{x^4}{24} +\cdots \label{eq_ps4}\tag{8.6.3} \end{equation}

We now find the derivative: \begin{align*} \fp(x) \amp = \infser n\frac{x^{n-1}}{n!}\\ \amp =\infser \frac{x^{n-1}}{(n-1)!} = 1+x+\frac{x^2}{2!}+\cdots.\\ \end{align*} Since the series starts at \(n=1\) and each term refers to \((n-1)\text{,}\) we can re-index the series starting with \(n=0\text{:}\) \begin{align*} \amp = \infser[0] \frac{x^{n}}{n!}\\ \amp = f(x). \end{align*}

We found the derivative of \(f(x)\) is \(f(x)\text{.}\) The only functions for which this is true are of the form \(y=ce^x\) for some constant \(c\text{.}\) As \(f(0) = 1\) (see Equation (8.6.3)), \(c\) must be 1. Therefore we conclude that \begin{equation*} f(x) = \infser[0] \frac{x^n}{n!} = e^x \end{equation*} for all \(x\text{.}\)

We can also find \(\ds \int f(x)\ dx\text{:}\) \begin{align*} \int f(x)\ dx \amp = C+\infser[0] \frac{x^{n+1}}{n!(n+1)}\\ \amp = C+ \infser[0] \frac{x^{n+1}}{(n+1)!} \end{align*}

We write out a few terms of this last series: \begin{equation*} C+ \infser[0] \frac{x^{n+1}}{(n+1)!} = C+ x+ \frac{x^2}2+\frac{x^3}{6}+\frac{x^4}{24}+\cdots \end{equation*}

The integral of \(f(x)\) differs from \(f(x)\) only by a constant, again indicating that \(f(x) = e^x\text{.}\)

The differential equation \(y' = 2y\) describes a function \(y=f(x)\) where the derivative of \(y\) is twice \(y\) and \(y(0)=1\text{.}\) This is a rather simple differential equation; with a bit of thought one should realize that if \(y=Ce^{2x}\text{,}\) then \(y' = 2Ce^{2x}\text{,}\) and hence \(y' = 2y\text{.}\) By letting \(C=1\) we satisfy the initial condition of \(y(0)=1\text{.}\)

Let's ignore the fact that we already know the solution and find a power series function that satisfies the equation. The solution we seek will have the form \begin{equation*} f(x) = \infser[0] a_nx^n = a_0+a_1x+a_2x^2+a_3x^3+\cdots \end{equation*} for unknown coefficients \(a_n\text{.}\) We can find \(\fp(x)\) using Theorem 8.6.7: \begin{equation*} \fp(x) = \infser a_n\cdot n\cdot x^{n-1} = a_1+2a_2x+3a_3x^2+4a_4x^3\cdots. \end{equation*}

Since \(\fp(x) = 2f(x)\text{,}\) we have \begin{align*} a_1+2a_2x+3a_3x^2+4a_4x^3\cdots \amp = 2\big(a_0+a_1x+a_2x^2+a_3x^3+\cdots\big)\\ \amp =2a_0+2a_1x+2a_2x^2+2a_3x^3+\cdots \end{align*}

The coefficients of like powers of \(x\) must be equal, so we find that \begin{equation*} a_1 = 2a_0, 2a_2 = 2a_1, 3a_3 = 2a_2, 4a_4 = 2a_3, \text{ etc. } \end{equation*}

The initial condition \(y(0) = f(0) = 1\) indicates that \(a_0 = 1\text{;}\) with this, we can find the values of the other coefficients: \begin{align*} a_0 = 1 \text{ and } a_1=2a_0 \amp \Rightarrow a_1 = 2;\\ a_1 = 2 \text{ and } 2a_2 = 2a_1 \amp \Rightarrow a_2=4/2 =2;\\ a_2=2 \text{ and } 3a_3 = 2a_2 \amp \Rightarrow a_3=8/(2\cdot3)=4/3;\\ a_3=4/3 \text{ and } 4a_4 = 2a_3 \amp \Rightarrow a_4 =16/(2\cdot3\cdot4)= 2/3. \end{align*}

Thus the first 5 terms of the power series solution to the differential equation \(y'=2y\) is \begin{equation*} f(x) = 1+ 2x+2x^2 + \frac43x^3+\frac23x^4+\cdots \end{equation*}

In Section 8.8, as we study Taylor Series, we will learn how to recognize this series as describing \(y=e^{2x}\text{.}\)