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?

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- Apr 8th 2009, 07:12 PMmollymcf2009Power 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? - Apr 8th 2009, 07:18 PMJhevon
- Apr 8th 2009, 07:36 PMmollymcf2009
- Apr 8th 2009, 07:38 PMmollymcf2009
- Apr 8th 2009, 07:54 PMChris L T521
- Apr 8th 2009, 08:01 PMmollymcf2009
- Apr 8th 2009, 08:08 PMJhevon
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. - Apr 8th 2009, 08:45 PMmatheagle
Do you mean n=1 instead of n+1?