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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.

northcity4
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1 Answers1

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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.

Batzorig Zorigoo
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  • "Then $f$ is continuous if its differentiable." - you shifted from proving $f$ is continuous on $(-1,1)$ to proving it's differentiable there (which is stronger statement) without explaining how to prove it. – Abstraction Dec 13 '16 at 11:03
  • @Batzorig_Zorigoo Let me see if this logic works: When $x=1,-1$ the series converges (proof by the alternating harmonic series test). By Abel's Theorem, it converges uniformly on the set $[0,1]$ and $[-1,0]$ respectively. Thus, it would converge uniformly on $[-1,1]$. (I figured why not show uniform convergence as it might help). Still lost on showing a series is continuous though (or disproving). – northcity4 Dec 14 '16 at 06:06
  • @Abstraction Editted. It should work. It is differentiable when the derivative is determined at that point. So I don't think proving $f$s differentiability in that region is that much hard work. – Batzorig Zorigoo Dec 14 '16 at 07:37