Thread: Power Series w/ factorial

Can someone check to see if I evaluated this correctly?

$\displaystyle \sum^{\infty}_{n+1} \frac{n!x^n}{8\cdot 17\cdot 26\cdot ...(9n-1)}$

$\displaystyle = \sum^{\infty}_{n+1} \frac{n!x^n}{(9n-1)!}$

Is that right?

Originally Posted by mollymcf2009

Can someone check to see if I evaluated this correctly?

$\displaystyle \sum^{\infty}_{n+1} \frac{n!x^n}{8\cdot 17\cdot 26\cdot ...(9n-1)}$

$\displaystyle = \sum^{\infty}_{n+1} \frac{n!x^n}{(9n-1)!}$

Is that right?

no

(9n - 1)! = (9n - 1)(9n - 2)(9n - 3)....(3)(2)(1)

Originally Posted by Jhevon

no

(9n - 1)! = (9n - 1)(9n - 2)(9n - 3)....(3)(2)(1)

Ok, I need a hint....

Originally Posted by Jhevon

no

(9n - 1)! = (9n - 1)(9n - 2)(9n - 3)....(3)(2)(1)

Ok, how about...9n!-1

Originally Posted by mollymcf2009

Ok, how about...9n!-1

It doesn't work for values of $\displaystyle n\geqslant3$!! XD

What in particular are you asked to do with this series? Show that it converges??

Originally Posted by Chris L T521

It doesn't work for values of $\displaystyle n\geqslant3$!! XD

What in particular are you asked to do with this series? Show that it converges??

Find the radius of convergence and the interval of convergence.

I don't know a good way to rewrite factorials. Do I need to do that before I try to evaluate this? Thanks Chris!

Originally Posted by mollymcf2009

Find the radius of convergence and the interval of convergence.

I don't know a good way to rewrite factorials. Do I need to do that before I try to evaluate this? Thanks Chris!

if that's the case, why not use the ratio or root test? you don't need to rewrite anything. probably the ratio test will be easier here

however, in general, you can use Stirling's approximation.

Do you mean n=1 instead of n+1?