interval of convergence of power series

**rcmango** · February 11th 2008, 03:53 PM

heres the problem: http://img211.imageshack.us/img211/4962/93204887ca3.png

i used the ratio test to get 4n+6 |x|

normally i would just subtract or add whats inside the absolute value to each side of x. and then divide by the 6|x| i believe.

however,
i'm really confused by the 4n+6, how do i get the radius of convergence now?
also, its possible this problem diverges.

thanks for any help.

**Soroban** · February 11th 2008, 06:09 PM

Hello, rcmango!

That function doesn't look right . . . $\sum^{\infty}_{n=1}\frac{(2n!)}{n!}\,x^n$
We can cancel the $n!$ and have: . $\sum^{\infty}_{n=1} 2x^n$

So I suspect that it is: . $\sum^{\infty}_{n=1}\frac{(2n)!}{n!}x^n$

The ratio is: . $\frac{[2(n+1)]!x^{n+1}}{(n+1)!} \cdot\frac{n!}{(2n)!x^n} \;=\;\frac{[2(n+1)]!}{(2n)!}\cdot\frac{n!}{(n+1)!} \cdot\frac{x^{n+1}}{x^n}$

. . $= \;\frac{2(n+1)(2n+1)}{1}\cdot\frac{1}{n+1}\cdot\fr ac{x}{1} \;=\;2(2n+1)x$ . . . which diverges.

**rcmango** · February 11th 2008, 06:26 PM

Okay, which brings me to my question again, its okay just to say it diverges at this point?

what is going on with the 2n+6 is it because the 2n? and i don't need to show any interval?

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