# Evaluating $\lim_{n\to \infty}\left(^nC_{0}\,^nC_{1}\cdots\,^nC_{n-1}\,^nC_{n}\right)^{1/(n(n+1))}$. Where's my error?

$$L=\lim_{n\to \infty}\left(\;^nC_{0}\,\cdot\,^nC_{1} \,\cdot\,\cdots\,\cdot\,^nC_{n-1}\,\cdot\,^nC_{n}\;\right)^{1/(n(n+1))}$$

My attmept:

Taking log on both sides:

$$\ln L= \lim_{n\to \infty} \frac{\ln (^nC_{0}) +\ln(^nC_{1})+\cdots+\ln(^nC_{n-1})+\ln(^nC_{n})}{n(n+1)}$$

Applying L'hospital rule (differentiating numerator and denominator with respect to $$n$$) because it is $$\dfrac{\infty}{\infty}$$ form

$$\ln L=\lim_{n\to \infty}\frac{0+\frac{1}{n}+\frac{2!(2n-1)}{n(n-1)}+\cdots+\frac{1}{n}+0}{2n+1}=\frac{0}{\infty}=0 \implies L=1$$

But answer is coming $$L= \sqrt{e}$$

i don't see where i'm doing wrong. in know how answer is coming $$\sqrt{e}$$ but

why L hospital rule isn't working .

i think there is no problem in differentiating function in numerator becuase all functions are continuos at $$n =\infty$$

• I've removed the precal tag since this is a calculus question. – GNUSupporter 8964民主女神 地下教會 Oct 1 '18 at 19:45
• A simpler version of the same idea: n times x is the same as adding x to itself n times. Since the derivative of x, with respect to x, is 1, the derivative of nx is the sum n 1's or n. That is true! But if we argue $x^2= x\cdot x$ is "x added to itself x times" so its derivative is just 1 add x times, x, then we get the wrong answer. The difficulty is that "adding it to itself x times" is itself a function of x where adding n times is not. – user247327 Oct 1 '18 at 20:01
• To solve the sum write $^nC_k = \frac{n!}{k!(n-k)!}$ to get $\log L_n = \frac{(n+1)\log(n!) - 2\sum_{k=0}^n \log(k!)}{n(n+1)}$. The latter sum can be simplified to a sum on the form $\sum k \log(k)$ and this can be estimated using an integral – Winther Oct 1 '18 at 21:00

Each term in your numerator $$\to 0$$, but the number of such terms is $$n+1$$, which $$\to\infty$$. So it's not obvious, for example, that their sum tends to $$0$$; indeed, it doesn't. If you want to solve the problem properly, I recommend the Stirling approximation $$k!\approx\sqrt{2\pi}k^{k+1/2}e^{-k}$$.
• On the recommendation: I don't see how Stirlings approximation is that useful here since $^nC_k = \frac{n!}{k!(n-k)!}$ and only $n$ is guaranteed to be large in the sum. – Winther Oct 1 '18 at 20:59
• @Winther Even for small $k$, the fractional error approximates $1/(12k)$. You can make it work with care. – J.G. Oct 1 '18 at 21:06