# Infinite Series with factorial

I'm having trouble manipulating the function of this series which has factorials to show that it converges or diverges using the ratio test.

The series is $\sum\limits_{k=1}^{\infty}\dfrac{(k!)^2}{(2k)!}$. The following are the steps I used, but I made a mistake expanding/canceling the factorial portion somewhere which I can't figure out..

First write the limit using ratio test: $$\lim\limits_{k\to\infty}\dfrac{((k+1)!)^2}{(2k+2)!}\cdot\dfrac{(2k)!}{(k!)^2}$$ Expand to cancel factorials: $$\lim\limits_{k\to\infty}\dfrac{((k+1)(k!))^2}{(2k+2)(2k+1){(2k)}!}\cdot\dfrac{(2k)!}{(k!)^2}$$

Then after canceling I got:

$$\lim\limits_{k\to\infty}\dfrac{(k+1)^2}{(2k+2)(2k+1)}$$

This would result in polynomials of the same degree in the numerator and denominator, and the coefficients of the $k^2$ term would be 1 and 4, so I said the series converges because $\dfrac{1}{4} < 1$ which means the series converges by the ratio test. However, the series is supposed to diverge according to my book. I don't have much practice working with factorials so I know my mistake must be somewhere in there. Any help is appreciated.

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Your work is correct. Seemingly your text is incorrect. – Cameron Williams Apr 2 '14 at 4:28
Wolfram Alpha confirms convergence: wolframalpha.com/input/…*+%28‌​i%21%29%2F%282i%29%21 – ml0105 Apr 2 '14 at 4:33
The work above is correct. The series is convergent. – kmitov Apr 2 '14 at 4:34
Yep, looks good to me as well. – MPW Apr 2 '14 at 4:39
$\binom{2k}{k}\sim\frac{4^k}{\sqrt{\pi k}}$. Thus, the terms of your series are $\sim\frac{\sqrt{\pi k}}{4^k}$ which shows why the ratio test gives $\frac14$. – robjohn Apr 2 '14 at 5:02

For integers $1 \leq j \leq k,$ we find $(j + k) \geq 2 j.$ As a result, $$\frac{(k!)^2}{(2k)!} = \prod_{j=1}^k \frac{j}{j+k} \leq \prod_{j=1}^k \frac{1}{2} = \frac{1}{2^k}$$
Detail, $$\frac{j+k}{j} = 1 + \frac{k}{j} \geq 1 + 1 = 2,$$ so $$\frac{j}{j+k} \leq \frac{1}{2}$$
Unless that book is the Bible, I suggest you take any information contained therein with a grain of salt. :-) As already noted by the other commenters, the series converges, so you did alright. More to the point, its value is $\displaystyle\sum_{n=1}^\infty\frac1{2n\choose n}=\frac13+\frac{2\pi\sqrt3}{27}$ . We know this because of the more general result $$\sum_{n=1}^\infty\frac{(2x)^{2n}}{{2n\choose n}n^2}=2\arcsin^2x\quad\iff\quad\Big(1-x^2\Big)\cdot\sum_{n=0}^\infty\frac{(2x)^{2n}}{2n\choose n}=1+\frac{x}{\sqrt{1-x^2}}\arcsin x,$$ $\quad$ which can be deduced by integrating the binomial series expansion of $\dfrac1{\sqrt{1-x^2}}=\arcsin'x$.