Find a convergence interval for the following power series
$\displaystyle ((-1)^n(x-2)^n)/3^2n$with the sigma symbol with infinity above and n=0 below.
(I appologize about the notation)
In General
For (-1)^n cn (x-a)^n
The ratio test yields |x-a|lim{c(n+1)/cn] < 1 where (n+1) is a subscript
|x-a| < lm{cn/c(n+1}
lm{cn/c(n+1} = R is called the radius of convergence
This guarantees convergence on the interval
|x-a| < R 0r (a-R < x < a+R)
Then check x =a-R and a +R in the original series to finish the interval of convergence
In your case cn = 1/3^(2n) ????? or 1/9n ?
If 1/3^(2n) Then R = 9 you have convergence on (-7,11) check the endpoints
Hello, mathatlast!
Find the convergence interval for: .$\displaystyle \sum^{\infty}_{n=0}\,(\text{-}1)^n\frac{(x-2)^n}{3^{2n}}$
$\displaystyle R \;=\;\left|\frac{a_{n+1}}{a_n}\right| \;=\;\left|\frac{(x-2)^{n+1}}{3^{2n+2}} \cdot\frac{3^{2n}}{(x-2)^n}\right| \;=\;\left|\frac{3^{2n}}{3^{2n+2}}\cdot \frac{(x-2)^{n+1}}{(x-2)^n}\right| \;=\;\left|\frac{1}{9}\,(x-2)\right|$
Then: .$\displaystyle \left|\frac{x-2}{9}\right| \:< \:1 \quad\Rightarrow\quad -1 \:<\:\frac{x-2}{9} \:<\:1 \quad\Rightarrow\quad -9 \:<\: x-2 \:<\:9 $
. . Therefore: .$\displaystyle -7 \:<\:x \:<\:11$
Um ... too slow again.
.
Thanks- I get it all, with the exception of where the 1/9 came about from 3^2n/3^2n+1....at least that is where I think the 1/9 came from. Can you explain it a little? I am sorry if it is obvious, but it has been a long day into night and my brain is melting!! Thanks