For this question, could I ask for some help regarding getting the eventual value of that combinatorial that involve an extra term. As normally $$\ C(n, n/2) = \frac{(2^n)}{\sqrt {\frac{n \pi}{2}}} $$
So how can I proceed to do this? As I am unsure on how to use $\ (1 + \frac{x}{k})^k -> e^x $ in solving this question.
Write out the binomial coefficient $C(\cdot, \cdot)$ in terms of factorials and apply Stirling's approximation. \begin{align} \binom{n}{n/2 + \sqrt{n}} &= \frac{n!}{(n/2 - \sqrt{n})!(n/2+\sqrt{n})!} \\ &\sim \frac{1}{\sqrt{2\pi}} \frac{n^{n+1/2}}{ (n/2- \sqrt{n})^{\frac{n}{2} - \sqrt{n} + \frac{1}{2}} (n/2+ \sqrt{n})^{\frac{n}{2} +\sqrt{n} + \frac{1}{2}} } \\ &= \frac{2^{n+1}}{\sqrt{2\pi n}}\frac{1}{ (1 - \frac{2}{\sqrt{n}})^{\frac{n}{2} - \sqrt{n} + \frac{1}{2}} (1 + \frac{2}{\sqrt{n}})^{\frac{n}{2} + \sqrt{n} + \frac{1}{2}} } \end{align}
When you multiply by $\sqrt{n}/2^n$, the factor in front becomes $\sqrt{2/\pi}$, and it remains to take the limit of the complicated second term.
The denominator can be rewritten as $$[(1-2/\sqrt{n})(1+2/\sqrt{n})]^{\frac{n}{2} - \sqrt{n} + \frac{1}{2}} \cdot (1+2/\sqrt{n})^{2\sqrt{n}}$$
Note that the first term is $(1-4/n)^{\frac{n}{2} - \sqrt{n} + \frac{1}{2}}$. By a hand-wavy argument, $(1-4/n)^n \approx e^{-4}$ for large $n$, so the above term approaches $e^{-2}$. (This can be made more rigorous.)
For the second term, $(1+2/\sqrt{n})^{\sqrt{n}} \to e^2$, so $(1+2/\sqrt{n})^{2\sqrt{n}} \to e^4$.