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Is there any formula for ${(1-\frac{1}{2^n})}^{2^n}$ ?

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up vote 3 down vote accepted

Using well-known Binomial Theorem, $(a+b)^m=\sum_{0\le r\le m}\binom m ra^{m-r}b^r$ where $m$ is a positive integer.

Put $a=1,b=-\frac1{2^n},m =2^n$ we shall get the required expansion.

Now we know from this, $$\lim_{m\to \infty}\left(1+\frac1m\right)^m=e\implies \lim_{m\to \infty}\left(1+\frac1m\right)^{mn}=e^n$$

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The notation as it is, As far as I know, has no simpler expression.

However, note that,

$$\lim_{n\rightarrow\infty}{\left(1-\frac{1}{2^n}\right)}^{2^n}=\frac{1}{e}$$ which you can prove.

Added : It is not so difficult to prove it. Just substitute $x=2^n$ and note that $x\rightarrow \infty $ as $n\rightarrow \infty$. Also note that: $$\lim_{x\rightarrow\infty}\left(1+\frac{(-1)}{x}\right)^x=e^{-1}$$

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Can you please write the main idea behind the prove? – Yoni Hassin Jan 7 '13 at 16:04

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