# Does the series $\sum_{n=1}^\infty\frac{(2n)!}{2^{2n}(n!)^2}$ converge or diverge.

$$\sum_{n=1}^\infty\frac{(2n)!}{2^{2n}(n!)^2}$$

Can I have a hint for whether this series converges or diverges using the comparison tests (direct and limit) or the integral test or the ratio test?

I tried using the ratio test but it failed because I got 1 as the ratio. The integral test seems impossible to use here.

• Your sum does not converge! – Dr. Sonnhard Graubner Oct 19 '18 at 12:54
• @Dr.SonnhardGraubner That doesn't seem to be a very helpful hint or answer... – DonAntonio Oct 19 '18 at 13:00
• Hint: Use Stirling's approximation on the factorials – Ronald Blaak Oct 19 '18 at 13:10
• Ronald Blaak this was what I was discussing with my friends but since we haven't been taught it, I doubt we are allowed to use it. – BuluBestTapu Oct 19 '18 at 13:24

For a more direct approach, you might directly expand the terms as follows: \begin{align} \frac{(2n)!}{4^n (n!)^2} &= \frac{1}{4^n}\frac{2n(2n-1)}{n^2}\frac{(2n-2)(2n-3)}{(n-1)^2}\cdots\frac{(4)(3)}{2^2} \frac{(2)(1)}{1^2} \\ &= \frac{2^n}{4^n}\frac{2n-1}{n} \frac{2n-3}{n-1}\cdots\frac{3}{2} \frac{1}{1} \\ &= \frac{4^n}{4^n} \frac{n - 1/2}{n} \frac{n - 3/2}{n-1} \cdots \frac{3/2}{2} \frac{1/2}{1}. \end{align} This is almost a telescoping product. By subtracting $$1/2$$ from each numerator (except the last), we get a smaller term that does telescope. Thus $$\frac{(2n)!}{4^n (n!)^2} \geq \frac{n-1}{n}\frac{n-2}{n-1} \cdots \frac{1}{2} \cdot (1/2) = \frac{1}{2n}.$$ Thus the $$n$$th term of your series is bigger than $$1/2n$$, and diverges by comparison with the harmonic series $$\sum_{n \geq 1} \frac{1}{n}.$$

• As a side comment --- when I checked this for myself, I also used Stirling's formula. But I know that I didn't know Stirling's bound when I first learned about series tests, so I understand the want for an answer without machinery. – davidlowryduda Oct 19 '18 at 13:55
• Very nice answer to a rather difficult (undergraduate level without Stirling's Formula) question. +1 – DonAntonio Oct 19 '18 at 14:17

Every term of the series is $$\frac{(2n)!}{2^{2n}(n!)^2}=\dfrac{\Gamma(2n+1)}{2^{2n}\Gamma^2(n)}=\dfrac{\Gamma(n+\frac12)}{n\Gamma(n)\sqrt{\pi}}>\dfrac{1}{2n}$$ by the formula $$\dfrac{\Gamma(n)}{\Gamma(2n)}=\dfrac{\sqrt{\pi}}{2^{2n-1}\Gamma(n+\frac12)}$$

• Note that the Gamma function is increasing. – Nosrati Oct 19 '18 at 13:32
• the moderators vote each other and leave people un-voted, it is ridiculous :) – Nosrati Oct 19 '18 at 14:00
• I can't say what the moderators do, and I don't really care, though I think they are not a closed sect and "vote for each other". Your answer uses Gamma Function and some of its properties, which seems to be (usually) stuff from middle undergraduate level and up. Perhaps that's why your answer hasn't been upvoted yet. – DonAntonio Oct 19 '18 at 14:21
• Lol. Btw you don't need votes, your answers are excellent. – manooooh Oct 19 '18 at 14:21

HINT

We have that

$$\frac{(2n)!}{2^{2n}(n!)^2} =\frac1{4^n}\binom{2n}{n} \sim \frac{1}{\sqrt{2\pi n}}$$

indeed recall that by bounds and asymptotic formulas for the binomial coefficient

$$\binom{2n}{n} \sim \frac{4^n}{\sqrt{2\pi n}}$$

then we can refer to limit comparison test.

Refer also to the related