Evaluating $\lim\limits_{n\to \infty }\sum\limits_{k=0}^n\:(2n)^{-k}\binom{n}{k}$

Question

$$\lim _{n\to \infty }\sum _{k=0}^n\:\frac{\binom{n}{k}}{\left(2n\right)^k}$$ I've got to the form: $$\lim _{n\to \infty }\frac{2^n\left(2n-1\right)}{\left(2n\right)^{n+1}-1}=\lim _{n\to \infty }\frac{2^{n+1}n-2^n}{2^{n+1}n^{n+1}-1}$$

And it should be $e^{1/2}$ but I always get $0$. I know it's just easy but I don't get it.

Answer 1

First use the binomial theorem:

$$\sum_{k=0}^n\binom{n}k\left(\frac1{2n}\right)^k=\left(1+\frac1{2n}\right)^n\;.$$

Now

$$\lim_{n\to\infty}\left(1+\frac1{2n}\right)^n=\lim_{n\to\infty}\left(\left(1+\frac1{2n}\right)^{2n}\right)^{1/2}=\left(\lim_{n\to\infty}\left(1+\frac1{2n}\right)^{2n}\right)^{1/2}\;,$$

and you should know what the last limit there is.