Question

I have this power series I need help with. I need to find its radius, interval of convergence. Then for what values of x, if any, does it converge absolutely? for what values of x does it converge conditionally? Not sure how to approach

Answer 1

\[\sum_{n=1}^{\infty} ((n!)^2/2^n(2n)!)x^n\]

Answer 2

sorry ifs its confusing.. That x^n is supposed to be outside of the fraction

Answer 3

Any real answers

Answer 4

\[\huge{\sum_{n=1}^{\infty}\frac{(n!)^2}{2^n(2n!)}x^n}\]is this it?

Answer 5

yes

Answer 6

It converges because 2^n.

Answer 7

yeah that doesnt help me.. not even sure how to find radi
us or anything

Answer 8

You have to do the analysis. x^n/2^n for x < 1 it converges absolutely. conditional for 1< x< 2. x > 2 it diverges.

Answer 9

hi paint

Answer 10

we find this by multiply the rule, or function of the series by its reciprocal modified by n-1

Answer 11

or if your more comfortable with an n+1 construction; divide it by its own reciprocal in n, but modify the first by n+1

either way, the construct is for user comfort, not for math comfort

Answer 12

\[\lim_{n\to\ inf}a_n*\frac{1}{a_{n-1}}\]
or

\[\lim_{n\to\ inf}a_{n+1}*\frac{1}{a_{n}}\]

Answer 13

Thanks Amistre, I sort of gave up on this and forgot to check back. I knew that way mostly, I just could was having problems simplifying I think. Appericate the help