1. ## [SOLVED] Power Series(Radius and interval of convergence)

hey there..
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..

2. Originally Posted by nameck
hey there..
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..

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

Tonio

3. 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?

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

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

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

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

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

6. 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?

7. 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 $x=-\frac{1}{8}$ you have $\frac{8^n}{n!}\left(-\frac{1}{8}\right)^n=\left(\frac{8}{-8}\right)^n\frac{1}{n!}=\frac{(-1)^n}{n!}$

is my radius of interval is correct?
No. The convergence radius is what VonNemo wrote and for the reasons he/she wrote.

Tonio