Thread: [SOLVED] Power Series(Radius and interval of convergence)

hey there..
i've try this question already...
i got radius of convergence, R = 1/8

at x = 1/8,
converges...

at x = -1/8
i don't how to do this one..
please help me..

Originally Posted by nameck

hey there..
i've try this question already...
i got radius of convergence, R = 1/8

at x = 1/8,
converges...

at x = -1/8
i don't how to do this one..
please help me..

At $\displaystyle x=-\frac{1}{8}$ you get $\displaystyle \sum\limits_{n=1}^\infty\frac{(-1)^n}{n!} = $ a Leibnitz series and thus convergent.

Tonio

the question is..
find the radius of convergence and interval of convergence of the
power series (attachment)

tonio.. is (8^n) / (-8) = -1^n?
is my radius of interval is correct?

$\displaystyle a_n=\frac{8^n}{n!}x^n$

$\displaystyle |\frac{a_{n+1}}{a_n}|=|\frac{8^{n+1}}{(n+1)!}x^{n+ 1}\cdot\frac{n!}{8^nx^n}|=\frac{8|x|}{n+1}$

Now, $\displaystyle \lim_{n\to\infty}\frac{8|x|}{n+1}=0<1$ $\displaystyle \forall{x}\in(-\infty,\infty)$.

Therefore, the series converges $\displaystyle \forall{x}$ by the ratio test.

This implies an infinte radius.

how did u get radius from infity to -infinity?
i haven't learn this kind solution(ration test to find radius and interval)

Originally Posted by nameck

how did u get radius from infity to -infinity?
i haven't learn this kind solution(ration test to find radius and interval)

Well, what tests for convergence do you know?

Originally Posted by nameck

the question is..
find the radius of convergence and interval of convergence of the
power series (attachment)

tonio.. is (8^n) / (-8) = -1^n?

No, and this is not what you have. At $\displaystyle x=-\frac{1}{8}$ you have $\displaystyle \frac{8^n}{n!}\left(-\frac{1}{8}\right)^n=\left(\frac{8}{-8}\right)^n\frac{1}{n!}=\frac{(-1)^n}{n!}$

is my radius of interval is correct?

No. The convergence radius is what VonNemo wrote and for the reasons he/she wrote.

Tonio