Exercise 6.5.1

(a)

$g(1)$ converges by the Alternating Series Test, so the radius of convergence is at least 1, and $g$ must be defined on at least $(-1,1]$. Theorem 6.5.1 and Abel’s Theorem together indicate that indicate that $g$ converges absolutely on $(-1,1]$ as well. Thus, since each term is continuous, $g(x)$ is continuous on $(-1,1]$.

$g$ is not defined at $-1$ since $g(-1)$ would otherwise be

$$\underset{n=1}{\overset{\infty }{\sum }}\frac{-1}{n}$$

which diverges.

$g$ cannot converge at any point $|x|>1$ because if it did, that would imply the radius of convergence is strictly larger than 1, and thus $g$ would need to converge at $-1$, which it doesn’t.

(b)

$g^{\prime }(x)$ is at least defined on $(-1,1)$, by Theorem 6.5.7, with the derivative given by

$$g^{\prime }(x)=\underset{n=0}{\overset{\infty }{\sum }}{(-x)}^n=\frac{1}{x+1}$$

$g^{\prime }(x)$ cannot be defined at $x\le -1$ since $g$ isn’t even defined there. To show that $g^{\prime }(1)$ is defined and is also given by this formula requires a bit more care, since the infinite sum does not actually converge for $1$. We return to the definition of the derivative:

$$\begin{array}{ccc}\hfill g^{\prime }(1)& =\underset{x\to 1}{\lim <!--\; FUNCTION\; APPLICATION\; -->}\frac{\underset{n=1}{\overset{\infty }{\sum }}\frac{{(-1)}^{n+1}}{n}-\underset{n=1}{\overset{\infty }{\sum }}\frac{{(-1)}^{n+1}}{n}{x}^n}{1-x}=\underset{x\to 1}{\lim <!--\; FUNCTION\; APPLICATION\; -->}\underset{n}{\overset{\infty }{\sum }}\frac{{(-1)}^{n+1}}{n}\frac{1-x^n}{1-x}\hfill & \hfill \\ \hfill & =\underset{x\to 1}{\lim <!--\; FUNCTION\; APPLICATION\; -->}\frac{1}{1-x}\underset{n}{\overset{\infty }{\sum }}\frac{{(-1)}^{n+1}}{n}\left(1-x^n\right)\hfill \end{array}$$

With some algebra, we can show that this converges by the alternating series test, keeping in mind that we can assume $x\in (0,1)$. We have $\frac{1-x^n}{n}<\frac{1}{n}\to 0$, so we just need to show $\frac{1-x^n}{n}\ge \frac{1-x^{n+1}}{n+1}$:

$$\begin{array}{ccc}\hfill \frac{1-x^n}{n}\ge \frac{1-x^{n+1}}{n+1}& ⟺(1-x^n)(n+1)\ge n-n{x}^{n+1}\hfill & \hfill \\ \hfill & ⟺n-n{x}^n+1-x^n\ge n-n{x}^{n+1}\hfill \\ \hfill & ⟺1-x^n\ge n{x}^n(1-x)\hfill \\ \hfill & ⟺\frac{1-x^n}{1-x}=\underset{i=0}{\overset{n-1}{\sum }}{x}^i\ge \underset{i=0}{\overset{n-1}{\sum }}{x}^n=n{x}^n\hfill \end{array}$$

Now we know that $g^{\prime }(1)$ exists. We can show that $g^{\prime }(1)=\frac{1}{1+1}=0.5$ by noting that $\frac{1}{x+1}$ is strictly decreasing on $[0,1)$, so in order for the derivative $g^{\prime }(x)$ to maintain the intermediate value property, $g^{\prime }(1)=0.5$.