Question

power series - radius of convergence
I started out with the series: 1+2x+4x^2/(2!)^2+6x^3/(3!)^2+...

so I did the ratio test for sum (0,∞) of (2n)!x^n/(n!)^2
and got
x lim n->∞ of 2/n+1
The answer is 1/4, but I don't understand how that's the answer. I obviously did something wrong because the radius of convergence I get is 0.. I think

Answer 1

(2n)!x^n/(n!)^2

^^^^
this is not the right representation of the sequence of terms

Answer 2

I still can't get 1/4 though, I must admit...

Answer 3

It is the right representation--I checked with my professor this afternoon.
\[\sum_{0}^{\infty}\frac{ (2n)!x^n }{ (n!)^2 }\]
I feel that my problem was I typed the sequence incorrectly (go figure).

\[1+2x+\frac{ 4!x^2 }{ (2!)^2 }+\frac{ 6!x^3 }{ (3!)^2 }+\frac{ 8!x^4 }{ (4!)^2 }+\frac{ 10!x^5 }{ (5!)^2 }+...\]

I did
\[\lim_{n \rightarrow \infty} \frac{ (2(n+1))!x^{n+1} }{ ((n+1)!)^2 }\times \frac{ ((n!)^2 }{ (2n)!x^n }\]

and canceled to get

\[x \lim_{n \rightarrow \infty} \frac{ 4n+2 }{ n+1 } \approx 4x\]

and \[4x < 1\]
so \[x < \frac{ 1 }{ 4 }\]
and the radius of convergence is \[\frac{ 1 }{ 4 }\]

Answer 4

Thank you for all your help though! I love that someone on here is familiar with calc 2 material.