Question

Find the radius of convergence and the interval of convergence of the series.

Series in comments

Answer 1

\[\sum_{k=1}^{\infty} \frac{ x^k }{ 2\times4\times6...(2k) }\]

Answer 2

looks like a good candidate for the ratio test

Answer 3

\[\frac{a_{n+1}}{{a_n}}=\frac{x}{2k+2}\] i think

Answer 4

did you just plug in k+1 and then cancel out with k?

Answer 5

essentially

\[\frac{ \cancel{2\times4\times6...(2k) }}{ \cancel{2\times4\times6...(2k)}~(2(k+1)) }=\frac{1}{2k+2}\]

Answer 6

So I would just find the limit after that?

Answer 7

yep, which can then be assessed for radius and interval convergences

Answer 8

|x|*0 <1 .... always

so the radius of convergence is: infinity, and the interval is (-inf,inf)