Infinite Series with factorial

Question

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.

Wolfram Alpha confirms convergence: http://www.wolframalpha.com/input/?i=sum%28i%3D0%2C+infinity%2C+%28i%21%29+*+%28i%21%29%2F%282i%29%21 — ml0105, Apr 02 '14 at 04:33
$\binom{2k}{k}\sim\frac{4^k}{\sqrt{\pi k}}$. Thus, the terms of your series are $\sim\frac{\sqrt{\pi k}}{4^k}$ which shows why the ratio test gives $\frac14$. — robjohn, Apr 02 '14 at 05:02
Thanks for the confirmation everyone. Although rare, I guess the text book can be wrong from time to time. ^.^ — Sabien, Apr 02 '14 at 05:44

score 5 · Answer 1 · answered Apr 02 '14 at 05:50

The series is supposed to diverge according to my book.

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

Will Jagy · Accepted Answer · 2014-04-02T06:10:30.943

1

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 $$ \frac{j}{j+k} \leq \frac{1}{2} $$

edited Apr 02 '14 at 06:10

answered Apr 02 '14 at 05:39

Will Jagy

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Infinite Series with factorial

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