What is the interval of convergence?
we proceed by the ratio test ( i hope you know it, if not, say so)
let An = (n^2 * x^2n)/2^2n
then by the ratio test, we want
lim{n-->infinity} |A(n+1)/An| < 1
lim{n-->infinity} |A(n+1)/An|
= lim{n-->infinity} |[(n+1)^2 * x^(2n + 2)]/2^(2n + 2) * (2^2n)/(n^2 * x^2n)|
= lim{n-->infinity} |(2^2n)/2^(2n + 2) * [(n+1)^2]/n^2 * x^(2n + 2)/x^2n|
= lim{n-->infinity} |2^-2 * (n^2 + 2n + 1)/n^2 * x|
= (1/4)|x|
for (1/4)|x| < 1
we must have |x| < 4
so -4 < x < 4
so our interval of convergence is (-4,4)
now we check the end points to see if that should be a closed interval
plug in -4 and then 4 for x and see if we have convergence
Note: E is summation
E (n^2 * (4)^2n)/2^2n = E (n^2 * 2^4n)/2^2n = E n^2 * 2^2n
this diverges as n--> infinity
a similar thing happens when x = -4
so the interval of convergence is (-4,4)
hmm, i got a different answer from TPH, wonder where i went wrong
oh, i see! i put x^(2n + 2)/x^2n = x (see the blue writing above), but it was supposed to be x^2. ok.
so just follow TPH's way, it's shorter--and correct
If you want to use the ratio test you need the fact that non of the terms are zero. That is, it fails for x=0. Hence you need to add "if x!=0 we can use the ratio test ....".
And then consider the case when x=0 (but that case is trivial).
I think you forgot a square somewhere.hmm, i got a different answer from TPH, wonder where i went wrong
You get that,
|x|<4
But it should be,
|x^2|<4