# power series, radius of convergence.

• January 25th 2008, 09:19 AM
rcmango
I've posted a few steps of the problem i solved in the pic: http://img180.imageshack.us/img180/9913/26377643by1.png

however, what is the radius of convergence? i got the interval to be [0, 1]

do i just solve for x for the radius? |(2x-1)|

thanks.
• January 25th 2008, 10:47 AM
Soroban
Hello, rcmango!

A small error in your algebra ... but you still got the right answer!

Quote:

$\sum^{\infty}_{n=1}\frac{(2x-1)^n}{n^4+16}$

$R \;=\;\left|\frac{a_{n+1}}{a_n}\right| \;=\;\left|\frac{(2x-1)^{n+1}}{(n+1)^4+16}\cdot\frac{n^4+16}{(2x-1)^n}\right| \;=\; \left|(2x-1)\cdot\frac{n^4+16}{(n+1)^4+16}\right|$

Then: . $\lim_{n\to\infty} R \;=\;\lim_{n\to\infty}\left|(2x-1)\cdot\frac{n^4+16}{(n+1)^4+16}\right| \;=\;{\color{blue}|2x-1|}$

Hence: . $|2x-1| \:<\:1\quad\Rightarrow\quad-1 \:<\:2x-1 \:<\:1\quad\Rightarrow\quad 0 \:<\:2x \:<\:2$

. . The interval of convergence is: . $0 \:<\:x \:<\:1$

. . And the radius of convergence is: . $r \:=\:\frac{1}{2}$

• January 26th 2008, 02:44 AM
rcmango
btw, the 0 and the 1 those both converge then correct? so there both on the closed interval [0, 1] right?

also, n^4 / (n+1)^4 does that just cancel?

thanks alot.