# Thread: interval of convergence

1. ## interval of convergence

sum n=o to oo (2n)!(x/2)^n

= (2n+2)!(x/2)^(n+1) / (2n)!(x/2)^n

= (2n+2)(2n+1)(x/2)

where do i go from there? ><

2. Originally Posted by jeph
sum n=o to oo (2n)!(x/2)^n

= (2n+2)!(x/2)^(n+1) / (2n)!(x/2)^n

= (2n+2)(2n+1)(x/2)

where do i go from there? ><
take the limit as n goes to infinity, and set whatever the result is to be < 1. then you find the x's that satisfy it and check the end points

3. Originally Posted by jeph
sum n=o to oo (2n)!(x/2)^n

= (2n+2)!(x/2)^(n+1) / (2n)!(x/2)^n

= (2n+2)(2n+1)(x/2)

where do i go from there? ><
Note,
$(2n)!(x/2)^n = \frac{(2n)!}{2^n}\cdot x^n$

Ratio test for $x\not =0$,
$\left| \frac{(2n+2)!}{2^{n+1}} \cdot \frac{2^n}{(2n)!} \cdot x\right| = \frac{(2n+2)(2n+1)}{2}\cdot |x|\to +\infty$

So, this convergense only for $x=0$

4. so does it go

lim n to oo [(2n+2)(2n+1)/2]x

= |[oo/2]x|<1

-1< [oo/2]x < 1
-2/oo< x < 2/oo
0< x <0

x=0 ?

5. Originally Posted by jeph
so does it go

lim n to oo [(2n+2)(2n+1)/2]x

= |[oo/2]x|<1

-1< [oo/2]x < 1
-2/oo< x < 2/oo
0< x <0

x=0 ?
Is there a number $x$ such that $0? No! But the thing is that we assumed (secretly) that $x\not = 0$. So the only point to be tested is $x=0$, which of course converges.