Pi sum

$\begin{aligned} L & = \lim_{n \to \infty} \frac {7\sqrt n}{4^n}{\color{#3D99F6}\sum_{k=0}^n {n \choose k}^2} & \small \color{#3D99F6} \text{See reference: } \sum_{k=0}^n {n \choose k}^2 = {2n \choose n} \\ & = \lim_{n \to \infty} \frac {7\sqrt n}{4^n}{\color{#3D99F6}{2n \choose n}} \\ & = \lim_{n \to \infty} \frac {7\sqrt n \color{#3D99F6}(2n)!}{4^n \color{#3D99F6}(n!)^2} & \small \color{#3D99F6} \text{By Stirling's formula: } n! \sim \sqrt{2\pi} n^{n+\frac 12}e^{-n} \\ & = \lim_{n \to \infty} \frac {7\sqrt n \cdot \color{#3D99F6}\sqrt{2\pi} (2n)^{2n+\frac 12}e^{-2n}}{4^n \cdot \color{#3D99F6}2\pi n^{2n+1}e^{-2n}} \\ & = \lim_{n \to \infty} \frac 7{\sqrt \pi} \end{aligned}$

$\implies a = \boxed{7}$

$$

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References:

• Equation 9: $\displaystyle \sum_{k=0}^n {n \choose k}^2 = {2n \choose n}$
• Stirling's formula