# Limit $\lim_{n\to \infty} \frac{(2n+3)!} {2^{n+2}\, (n+1)!\, (n+1)! }$

Is there a way to calculate the following limit:

$$\lim_{n\to \infty} \frac{(2n+3)!} {2^{n+2}\, (n+1)!\, (n+1)! }$$

I tried manipulating the numerator and forming an inequality to use squeeze theorem but couldn't get anything useful.

• Have you tried Stirling approximation? May 6, 2017 at 10:59

An idea: use the inverse of your sequence and put

$$a_n=\frac{2^{n+2}(n+1)!(n+1)!}{(2n+3)!}\implies \frac{a_{n+1}}{a_n}=\frac{2^{n+3}(n+2)!(n+2)!}{(2n+5)!}\cdot\frac{(2n+3)!}{2^{n+2}(n+1)!(n+1)!}=$$

$$=\frac{2(n+2)(n+2)}{(2n+4)(2n+5)}\xrightarrow[n\to\infty]{}\frac12<1$$

so by the quotient rule the series $\;\sum\limits_{n=1}^\infty a_n\;$ converges, which means

$$\lim_{n\to\infty} a_n=0\implies \lim_{n\to\infty}\frac1{a_n}=\infty$$

• =D Like minds think alike they do say. May 6, 2017 at 11:03
• @SimplyBeautifulArt Yes indeed...I only added the series flavour. May 6, 2017 at 11:04

$$a_n=\frac{(2n+3)!}{2^{n+2}[(n+1)!]^2}$$

By the ratio test we find that

$$\lim_{n\to\infty}\frac{a_{n+1}}{a_n}=\lim_{n\to\infty}\frac{(2n+5)(2n+4)}{2(n+2)^2}=2$$

And since $a_n>0$, we see that it diverges to $+\infty$ bounded by $k_12^n<a_n<k_22^n$ for some $k_1$ and $k_2$ greater than zero.

• @Astyx Unless you use this type of asymptote, but yeah, I suppose that is the case. May 6, 2017 at 11:31
• I guess you mean a Big Theta ($\Theta$) May 6, 2017 at 11:33
• @Astyx Which is equivalent to my so called asymptote for exponential functions :D May 6, 2017 at 11:34

Another solution using Stirling's approximation :

$$n! \sim \left({n\over e}\right)^n\sqrt{2\pi n}$$

Therefore \begin{align}{(2n+3)!\over 2^{n+2}(n+1)!^2} &\sim {\left({2n+3\over e}\right)^{2n+3}\sqrt{2\pi(2n+3)}\over 2^{n+2}\left({n+1\over e}\right)^{2(n+1)}2 \pi (n+1)}\\ &\sim {(n+3/2)^{2n+3} \over (n+1)^{2n+2}}{2^{n+2}\over e\sqrt{\pi n}}\\ &\sim \left(1+{1\over 2(n+1)}\right)^{2n+2} {2^{n+2}\over e}\sqrt{n\over \pi}\\ &\sim 2^{n+2}\sqrt{n\over \pi}\end{align}

Which clearly diverges to $\infty$.