If $n\in \mathbb{N},$ then $(1-1/2) (1-1/4)(1-1/8)(1-1/16)...(1-1/2^n)\geq 1/4+1/2^{n+1}.$

Question

I would like to know how to prove the following statement:

If $n\in \mathbb{N},$ then $$ \left(1-\frac{1}{2}\right) \left(1-\frac{1}{4}\right)\left(1-\frac{1}{8}\right)\left(1-\frac{1}{16}\right)...\left(1-\frac{1}{2^n}\right)\geq \frac{1}{4}+\frac{1}{2^{n+1}}.$$

This is my attempt: $$ \left(1-\frac{1}{2}\right) \left(1-\frac{1}{4}\right)\left(1-\frac{1}{8}\right)\left(1-\frac{1}{16}\right)...\left(1-\frac{1}{2^{n+1}}\right)\geq \left( \frac{1}{4}+\frac{1}{2^{n+1}}\right)\left(1-\frac{1}{2^{n+1}}\right)$$ $$=\frac{1}{4}+\frac{.}{.}$$

Thanks

Masik

Answer 1

Hint: How can you get from $$\left (1-\frac{1}{2}\right)\left (1-\frac{1}{4}\right)\left (1-\frac{1}{8}\right)\dots \left (1-\frac{1}{2^n}\right)$$ to $$\left (1-\frac{1}{2}\right)\left (1-\frac{1}{4}\right)\left (1-\frac{1}{8}\right)\dots \left (1-\frac{1}{2^{n+1}}\right)$$?

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From the comment I see that you have thought about this, which is great, and you should include your thoughts in your question. That way it's less probable that your post will be closed. You have however made a small error. Notice that:

$$\left (\frac{1}{4}+\frac{1}{2^{n+1}}\right)\left (1-\frac{1}{2^{n+1}}\right)=\frac{1}{4}-\frac{1}{2^{n+3}}+\frac{1}{2^{n+1}}-\frac{1}{2^{2n+2}}$$ so you only have to check that the inequality $$\frac{1}{2^{n+1}}-\frac{1}{2^{2n+2}}-\frac{1}{2^{n+3}}\geq \frac{1}{2^{n+2}}$$ holds. If you multiply boths sides by $2^{n+2}$ you get $$2-\frac{1}{2^n}-\frac{1}{2}\geq 1$$

Answer 2

Note that$$\frac{\left(1-\frac{1}{2}\right) \left(1-\frac{1}{4}\right)\left(1-\frac{1}{8}\right)\left(1-\frac{1}{16}\right)\cdots\left(1-\frac{1}{2^{n+1}}\right)}{\left(1-\frac{1}{2}\right) \left(1-\frac{1}{4}\right)\left(1-\frac{1}{8}\right)\left(1-\frac{1}{16}\right)\cdots\left(1-\frac{1}{2^n}\right)}=1-\frac1{2^{n+1}}.$$Since both expressions that you have are equal to $\frac12$ when $n=1$, all you need to prove is that$$(\forall n\in\mathbb{N}):1-\frac1{2^{n+1}}\geqslant\frac{\frac14-\frac1{2^{n+2}}}{\frac14-\frac1{2^{n+1}}}.$$Can you take it from here?

Answer 3

You can use Bernoulli's inequality: \begin{align} \Bigl(1-\frac12\Bigr)\Bigl(1-\frac14\Bigr)\dots\Bigl(1-\frac1{2^n}\Bigr)&=\frac12\Bigl(1-\frac14\Bigr)\dots\Bigl(1-\frac1{2^n}\Bigr)\\ &\ge \frac12\biggl[1- \Bigl(\frac14+\dots+\frac1{2^n}\Bigr)\biggr]=\frac12\biggl[1- \frac14\Bigl(1+\dots+\frac1{2^{n-2}}\Bigr)\biggr]\\ &=\frac12\biggl[1-\frac12\Bigl(1-\frac1{2^{n-1}}\Bigr)\biggr]=\frac12\biggl(\frac12+\frac1{2^n}\biggr). \end{align}