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$$\lim _{n\to \infty }\sum _{k=0}^n\:\frac{\binom{n}{k}}{\left(2n\right)^k}$$ 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.

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    How did you get to that form? That form seems wrong. – 2017-01-03
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    Title $\ne$ Body. – 2017-01-03
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    Use the [binomial theorem](https://en.wikipedia.org/wiki/Binomial_theorem#Theorem_statement) with $y = 1$ and $x=\frac{1}{2n}$. You should edit the question and fix the mismatch between the title and body. – 2017-01-03
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    @Winther Thank you ! I made a big mistake before getting to that form anyway... – 2017-01-03
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    $\lim _{n\to \infty }\sum _{k=0}^n\:\frac{\binom{n}{k}}{\left(2n\right)^k}=\sqrt{e}$, this is true. As for the second limit you wrote, it's $0$ – 2017-01-03

1 Answers 1

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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.