Find the radius of convergence and interval of convergence for the following summation:
$$\sum_{n=1}^{\infty}\frac{(2n)!}{n!}x^{3n}$$
I don't know how to deal with the $(2n)!$ in this question. Any help is appreciated!
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I'm assuming the x^3n in your title means $x^{3n}$.
Hint. Use the ratio test: $$\left|\frac{(2n+2)!}{(n+1)!}x^{3n+3}\bigg/\frac{(2n)!}{n!}x^{3n}\right| =\frac{(2n+2)(2n+1)(2n)!}{(n+1)n!}\frac{n!}{(2n)!}|x|^3$$ and now lots of things cancel.
David
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Thank you! Could you please explain how the (2n+2)! turns into (2n+2)(2n+1)(2n)! ? – user138069 Apr 01 '14 at 22:48
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$(2n+2)!=(2n+2)(2n+1)(2n)(2n-1)(2n-2)(2n-3)(2n-4)\cdots1$. – David Apr 01 '14 at 22:51
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Hint: $$\dfrac{(x+1)!}{x!}=\dfrac{(x+1)\cdot x!}{x!}=x+1$$