# Thread: how do you simplify this series and test for convergence?

1. ## how do you simplify this series and test for convergence? IMAGE-SOLUTION included..

I have the series Σ from k=1 to infinity, [(k!)^2]/(2k)!

I don't even know how to perform a ratio test cause I don't have any clue how i could simplify this... :/

2. ## Re: how do you simplify this series and test for convergence?

$\displaystyle \lim_{k \to \infty}\frac{(k!)^2/(2k)!}{(k+1!)^2/(2k+2)!}=\lim_{k \to \infty}\frac{(2k+1)(2k+2)}{(k+1)(k+1)}=4$

convergent

3. ## Re: how do you simplify this series and test for convergence?

Originally Posted by nappysnake
I have the series Σ from k=1 to infinity, [(k!)^2]/(2k)!
I don't even know how to perform a ratio test cause I don't have any clue how i could simplify this... :/
Consider the following:
\displaystyle \begin{align*}(2n)! &=(2n)(2n-1)(2n-2)\cdots(2)(1)\\ &=2^n(n)(n-1)\cdots(1)(2n-1)(2n-3)\cdots(1)\\ &=2^n(n!)(2n-1)(2n-3)\cdots(1) \end{align*}

Now $\displaystyle \frac{(n!)^2}{(2n)!}=\frac{n!}{2^n(2n-1)(2n-3)\cdots(3)(1)}$

4. ## Re: how do you simplify this series and test for convergence?

wnvl and Plato, thank you, you helped me come to some realizations!