$\lim_{n\rightarrow\infty}\sum_{k=1}^{n}{2n\choose k}\frac{1}{4^{n}}$ and $\lim_{n\rightarrow\infty}\sum_{k=1}^{2n}{2n\choose k}\frac{1}{4^{n}}$ is

Question

$\lim_{n\rightarrow\infty}\sum_{k=1}^{n}{2n\choose k}\frac{1}{4^{n}}=\lim_{n\rightarrow\infty}(1+\frac{1}{4^n})^{2n}$

using $(1+x)^n=1+nx+\frac{n(n-1)x^2}{2!}….$

$\lim_{n\rightarrow\infty}=1+\frac{2n}{4^n}+\frac{2n(2n-1)}{4^{2n}2!}….$

$\lim_{n\rightarrow\infty}=\frac{2n}{4^n}=\frac{\infty}{\infty}=\frac{2}{4^n\ln4}$

all terms vanish we are left with $1$ only.

Am I correct that limit is $1$?

Answer 1

If you want to find the limit of $ \sum\limits_{k=0}^{2n} \binom {2n} {k} \frac 1{4^{k}}$ then you can write the sum as $(1+\frac 1 {4})^{2n}=(\frac 5 4 )^{2n}$ and the limit is $\infty$

However, if you want to find the limit of $ \sum\limits_{k=0}^{2n} \binom {2n} {k} \frac 1{4^{n}}$ then the answer is obviously $1$ since the value is $1$ for every $n$.