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

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 $n =\infty$

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  • $\begingroup$ I've removed the precal tag since this is a calculus question. $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Oct 1 '18 at 19:45
  • $\begingroup$ 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. $\endgroup$ – user247327 Oct 1 '18 at 20:01
  • $\begingroup$ 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 $\endgroup$ – Winther Oct 1 '18 at 21:00
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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 $k!\approx\sqrt{2\pi}k^{k+1/2}e^{-k}$.

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  • $\begingroup$ +1 thank you that's very subtle point that sum of each terms which are tending to zero when added up infinitely .they may not be zero, infact here they are not ......i do agree with your logic but i will wait for some other answers too .......... $\endgroup$ – user454960 Oct 1 '18 at 20:10
  • $\begingroup$ 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. $\endgroup$ – Winther Oct 1 '18 at 20:59
  • $\begingroup$ @Winther Even for small $k$, the fractional error approximates $1/(12k)$. You can make it work with care. $\endgroup$ – J.G. Oct 1 '18 at 21:06

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