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.
Best Answer
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 $\dfrac1{\sqrt{1-x^2}}=\arcsin'x$.