Who's up to the challenge? 68

Since Stirling's approximation tells us that $\ln k! \; = \; (k+\tfrac12)\ln k - k + \ln\sqrt{2\pi} + O(k^{-1}) \qquad \qquad k \to \infty$ we can show that $2\sum_{k=0}^n \ln k! \; = \; (n+\tfrac32)\ln n! - \tfrac12n(n+1) + n\ln\sqrt{2\pi} + O(\ln n) \qquad \qquad n \to \infty$ so that $\begin{array}{rcl} \displaystyle \ln\left(\prod_{k=0}^n \binom{n}{k}\right) & = & \displaystyle (n+1) \ln n! - 2\sum_{k=0}^n \ln k! \\ & = & \displaystyle -\tfrac12 \ln n! + \frac12n(n+1) - n\ln \sqrt{2\pi} + O(\ln n) \\ & = & \displaystyle -\tfrac12 n \ln n + \tfrac12n(n+2) - n\ln\sqrt{2\pi} + O(\ln n) \qquad \qquad n \to \infty \end{array}$ But then $\ln\left(\sqrt[n]{\prod_{k=0}^n \binom{n}{k}}\right) \; = \; -\tfrac12\ln n + \tfrac12(n+2) - \ln\sqrt{2\pi} + O\big(\tfrac{\ln n}{n}\big) \qquad \qquad n \to \infty$ and hence $\ln\left(\frac{\sqrt{n} \sqrt[n]{\prod_{k=0}^n \binom{n}{k}}}{e^{\frac12n}}\right) \; = \;1 - \ln\sqrt{2\pi} + O\big(\tfrac{\ln n}{n}\big) \qquad \qquad n \to \infty$ Thus the desired limit is $\frac{e}{\sqrt{2\pi}}$ , making the answer $1+2=\boxed{3}$ .

Let $P_n = \displaystyle\prod_{k=1}^{n} \binom{n}{k}$ .

$P_n = \prod_{k=1}^{n-1} \binom{n}{k} = \prod_{k=1}^{n-1} \frac{n!}{k! (n-k)!} = \dfrac{(n!)^{n-1}}{\prod_{k=1}^{n} k^{2(n-k)}} = \dfrac{(n!)^{n-1}}{(n!)^{2n}} \prod_{k=1}^{n} k^{2k} = \dfrac{1}{(n!)^{n+1}} \left(\prod_{k=1}^{n} k^{k}\right)^2$

Now let $A$ represent the Glashier-Kinkelin constant. Then we can write an asymptotic expression for $P_n$ :

$P_n \approx \dfrac{1}{(n!)^{n+1}} \left(A n^{n^2/2+n/2+1/12}e^{-n^2/4}\right)^2 \approx \left(\dfrac{e^n} {n^n\sqrt{2\pi n}}\right)^{n+1} A^2 n^{n^2+n+1/6}e^{-n^2/2}$ $P_n \approx \dfrac{A^2 e^{n+n^2/2}}{(2\pi)^{n/2+1/2} n^{n/2+1/3}}$ $\sqrt[n]{P_n} \approx \dfrac{e^{1+n/2}}{\sqrt{2\pi n}}$

Therefore,

$\lim_{n\to\infty} \dfrac{\sqrt{n}\sqrt[n]{P_n}}{e^{n/2}} = \lim_{n\to\infty} \dfrac{\sqrt{n}}{e^{n/2}} \dfrac{e^{1+n/2}}{\sqrt{2\pi n}} = \boxed{\dfrac{e}{\sqrt{2\pi}}}$