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.
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.
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 $