Determine if the power series is continuous

Question

I am still getting a hold of power-series, so any help is appreciated. I am being to asked to determine if $g(x)$ is defined/converge/continuous on the following intervals: (I have also included relevant theorems I used)

$g(x)=\sum_{n=1}^{\infty} \frac{(-1)^{n+1}\cdot x^n}{n}$

$A=(-1,1)$

$B=(-1,1]$

$C=[-1,1]$

$Abel's Theorem$

$Theorem$ $6.5.1$: If a power series converges at some point $x_o\in \mathbb{R}$, then it converges absolutely for any $x$ satisfying $|x|<|x_o|$

Here is my attempt:

$g(x)$ is defined on this set. Further more, $g(x)$ converges at $1$ since when $x=1$, we get the alternating harmonic series (already proved this converges). By Theorem 6.5.1, $g(x)$ converges absolutely on $|x|<1$ which implies $-1<x<1 \Rightarrow$ converges on $A$

My question is if someone could help me along about how to show if it is continuous.

Answer 1

$g(x)=\sum_{n=1}^{\infty}\frac{(-1)^{n+1}x^n}{n}=-\sum_{n=1}^{\infty}\frac{(-1)^nx^n}{n}=-(-x+\frac{x^2}{2}-\frac{x^3}{3}+\frac{x^4}{4}+...)$

Consider $S=-1+x-x^2+x^3-x^4+...$. Clearly, if we were to integrate this with respect to $x$ and multiply it by $-1$, we'd have $g(x)$.

Now note that $xS=-x+x^2-x^3+x^4+...$

Adding them together yields $S(1+x)=-1$

Which means $S=-\frac{1}{1+x}$

In other words, $g(x)=-\int Sdx=-\int -\frac{1}{1+x}dx=\int \frac{1}{1+x}dx=\operatorname{ln}(1+x)+C$

The natural logarithm is convergent for all nonzero, positive inputs. In this case, the input is $1+x$, so we have $1+x\gt 0$ and therefore $x\gt -1$

Among the intervals presented, only $C$ includes $-1$. Therefore $g(x)$ is convergent on intervals $A$ and $B$.

Answer 2

You should use Alternating Series Test . (x^n)/n goes to infinity when | x | > 1. So by Alternating Series Test, your series diverges. And when x = 1, (x^n)/n goes to 0 as n goes to infinity. Using Alternating Series Test and Abel's theorem, we can prove that series converges when | x | <= 1. So the radius of convergence of your power series is 1.

You can't prove the radius of convergence is 1 using only Abel's theorem as it doesn't say anything when | x | > 1.

My question is if someone could help me along about how to show if it is continuous.

Let's say the series is a Tailor series of function f. Then f is continuous if its differentiable.

Edit: The function f should be something like x^-1 or x^-2 and take Tailor series with center x = 1. And f (x+1) should be your alternating series. You can easily take derivative and check its differentiability in the region you want.