Interval of Convergence Problem

Jan 2010
13
1
Sorry, but I have no idea how to use the math symbol thingies, so bear with me here.

How would I go about finding the interval of convergence of:

Σ (((-1)^n)(x^n))/n
Also, can the endpoints be included in the interval?
Thanks so much.
 

Prove It

MHF Helper
Aug 2008
12,883
4,999
Sorry, but I have no idea how to use the math symbol thingies, so bear with me here.

How would I go about finding the interval of convergence of:

Σ (((-1)^n)(x^n))/n
Also, can the endpoints be included in the interval?
Thanks so much.
You will need to use the Ratio Test.

Remembering that the Ratio Test tells us that the series is convergent where \(\displaystyle \lim_{n \to \infty}\left|\frac{a_{n + 1}}{a_n}\right|<1\), divergent where \(\displaystyle \lim_{n \to \infty}\left|\frac{a_{n + 1}}{a_n}\right|>1\) and inconclusive where \(\displaystyle \lim_{n \to \infty}\left|\frac{a_{n + 1}}{a_n}\right|=1\), no, you can not use the endpoints in your interval. However, there are other tests you can do for the endpoints.


Anyway, you have \(\displaystyle \sum_{n = 0}^{\infty}\frac{(-1)^nx^n}{n}\).

This will be convergent where

\(\displaystyle \lim_{n \to \infty}\left|\frac{a_{n + 1}}{a_n}\right|<1\)

\(\displaystyle \lim_{n \to \infty}\left|\frac{\frac{(-1)^{n + 1}x^{n + 1}}{n + 1}}{\frac{(-1)^nx^n}{n}}\right| < 1\)

\(\displaystyle \lim_{n \to \infty}\left|\frac{\frac{x^{n + 1}}{n + 1}}{\frac{x^n}{n}}\right| < 1\)

\(\displaystyle \lim_{n \to \infty}\left|\frac{nx^{n + 1}}{(n + 1)x^n}\right| < 1\)

\(\displaystyle \lim_{n \to \infty}\left|\frac{nx}{n + 1}\right| < 1\)

\(\displaystyle |x|\lim_{n \to \infty}\left(\frac{n}{n + 1}\right) < 1\)

\(\displaystyle |x| \lim_{n \to \infty}\left(1 - \frac{1}{n + 1}\right) < 1\)

\(\displaystyle |x| (1) < 1\)

\(\displaystyle |x| < 1\)

\(\displaystyle -1 < x < 1\).


So the series is convergent where \(\displaystyle -1 < x < 1\) and divergent where \(\displaystyle x < -1\) and \(\displaystyle x > 1\). You will need to use a different test for the endpoints of the interval.
 

skeeter

MHF Helper
Jun 2008
16,216
6,764
North Texas
Sorry, but I have no idea how to use the math symbol thingies, so bear with me here.

How would I go about finding the interval of convergence of:

Σ (((-1)^n)(x^n))/n
Also, can the endpoints be included in the interval?
Thanks so much.
set up the ratio test ...

\(\displaystyle \lim_{n \to \infty} \left|\frac{x^{n+1}}{n+1} \cdot \frac{n}{x^n} \right| < 1\)

\(\displaystyle |x| \lim_{n \to \infty} \frac{n}{n+1} < 1\)

\(\displaystyle |x| \cdot 1 < 1\)

\(\displaystyle -1 < x < 1\)

if \(\displaystyle x = -1\) , nth term is \(\displaystyle \frac{1}{n}\) ... the harmonic series diverges.

if \(\displaystyle x = 1\) , nth term is \(\displaystyle \frac{(-1)^n}{n}\) ... alternating harmonic series converges

interval of convergence is \(\displaystyle -1 < x \le 1 \)