URGENT: please help--power series/ratio test/roc

[chopchop]

I really need help with the following problem:

Use ratio test to find the radius of convergence of the power series

1+2x+(4!x^2)/(2!)^2 + (6!x^3)/(3!)^2+(8!x^4)/(4!)^2+(10!x^5)/(5!)^2+...

Answer supposed to be 1/4 (radius of conv.= 1/4)

So...this is what I have so far, but it's prob wrong

Cn=2n!/(n!)^2 ... (2n!X^n) / (n!)^2

so... abs val [ (An+1) / (An) ] =

[ 2(n+1)! / (n+1)!^2 ] X^(n+1) ]
_______________________________

[ 2n! / (n!) ^2] X^n

I know the the x's reduce, so there's just X on the top....then...

2!(n+1) (n!)^2
X times __________ times ______

(n+1)!^2 2!n



but then wouldn't (n+1) cancel, and then the two's, ...and I have not idea how the answer is 1/4--well here it should somehow be 4 and since it's the ratio, the radius of convergence would be 1/4

 

[pka]

\(\displaystyle \L
\frac{{\frac{{\left( {2k + 2} \right)!}}{{\left[ {\left( {k + 1} \right)!} \right]^2 }}}}{{\frac{{\left( {2k} \right)!}}{{\left[ {\left( k \right)!} \right]^2 }}}} = \frac{{\left( {2k + 2} \right)(2k + 1)}}{{\left[ {k + 1} \right]^2 }}\)

 

[chopchop]

I still don't know how to get the answer, I'm so brain-dead right now. Been working on this forever. Can you please help me further...how do you end up gettin 4?

 

[pka]

\(\displaystyle \L
\frac{{\left( {2k + 2} \right)(2k + 1)}}{{\left[ {k + 1} \right]^2 }} = \frac{{4k^2 + 6k + 2}}{{k^2 + 2k + 2}} \to 4\)

 

[chopchop]

Can you please tell me how, i just don't get it...feel stupid. thank you soooo much

 

[chopchop]

OOOps, sorry...i get it...I love you!!!!!!!

 

[pka]

\(\displaystyle \L
\begin{array}{l}
\lim _{k \to \infty } \frac{{4k^2 + 6k + 2}}{{k^2 + 2k + 2}} = \lim _{k \to \infty } \frac{{4 + 6/k + 2/k^2 }}{{1 + 2/k + 2/k^2 }} = 4 \\
4|x| < 1\quad \Rightarrow \quad |x| < 1/4 \\
\end{array}\)

 

[chopchop]

Tahnk u I got that... but how did u get [ (2n+2) (2n+1) ] / (n+1)^2