we can write $\frac{1}{4^n}$ as $\frac{1}{2^{2n}}$ then I used Stirling’s approximation for factorial for Simplification, but calculation became complicated, then I tried to use L' hopital's rule Rule again the calculation became complicated, Is there a different approach to this problem if so please do explain
Thanks
We have:
$$\sum_{k=0}^{n}\binom{2n}{k}\frac{1}{4^{n}} = \frac{1}{4^{n}}\sum_{k=0}^{n}\binom{2n}{k} = \frac{1}{4^{n}}\bigg(\frac{1}{2}\sum_{k=0}^{2n}\binom{2n}{k} + \frac{1}{2}\binom{2n}{n}\bigg)=\frac{1}{4^{n}}\bigg(\frac{4^{n}}{2} + \frac{1}{2}\binom{2n}{n}\bigg)$$
Using the approximation $\displaystyle\binom{2n}{n} \approx \frac{4^{n}}{\sqrt{\pi n}}$, derived from Stirling's Approximation:
$$\Longrightarrow\lim_{n\to\infty}\frac{1}{4^{n}}\bigg(\frac{4^{n}}{2} + \frac{1}{2}\binom{2n}{n}\bigg) = \lim_{n\to\infty}\frac{1}{4^{n}}\bigg(\frac{4^{n}}{2}+\frac{4^{n}}{2\sqrt{\pi n}}\bigg)=\lim_{n\to\infty}\bigg(\frac{1}{2}+\frac{1}{2\sqrt{\pi n}}\bigg)=\boxed{\frac{1}{2}}$$