Prove using induction $\prod_{i=1}^n (1-2^{-i}) ≥ \frac{1}{4} + 2^{-(n+1)}$ for all n∈N

Question

Prove using induction $$\prod_{i=1}^n (1-2^{-i}) ≥ \frac{1}{4} + 2^{-(n+1)}$$ for all n∈N

My answer so far:

base case: n = 1. P(1) = $$(1-2^{-1}) = \frac{1}{2} ≥ \frac{1}{4} + 2^{-(1+1)} $$ so it holds.

induction hypothesis: Assume P(k): $$\prod_{i=1}^k (1-2^{-i}) ≥ \frac{1}{4} + 2^{-(k+1)}$$

induction step:

$$\prod_{i=1}^{k+1} (1-2^{-i}) = (1-2^{-(k+1)})\cdot\prod_{i=1}^k (1-2^{-i}) $$

$$ ≥ (1-2^{-(k+1)})\cdot(\frac{1}{4} + 2^{-(k+1)})$$

I don't know where to go from here. I tried expanding to pull out the right inequality but can't seem to get it. Any help would be appreciated.

Answer 1

\begin{align} (1-2^{-(k+1)})(2^{-2}+2^{-(k+1)}) &= 2^{-2}+2^{-(k+1)}-2^{-2}2^{-(k+1)}-2^{-2(k+1)}\\ &= 2^{-2}+2^{-(k+1)}(1-2^{-2}-2^{-(k+1)})\\ &\geq 2^{-2}+2^{-(k+1)}(1-2^{-2}-2^{-2})\\ &= 2^{-2}+2^{-(k+2)} \end{align}

Answer 2

you just need to develop the product and the results comes by itself.

$$\prod_{i=1}^{k+1} (1-2^{-i}) = (1-2^{-(k+1)})\cdot\prod_{i=1}^k (1-2^{-i}) $$

$$ ≥ (1-2^{-(k+1)})\cdot(\frac{1}{4} + 2^{-(k+1)})$$

$$ ≥ \frac{1}{4} + \frac{1}{2^{k+1}}(1-(\frac{1}{4}+\frac{1}{2^{k+1}})) $$

and because $$ \forall k \in \mathbb{N}: \; \frac{1}{4}+\frac{1}{2^{k+1}} \leq \frac{1}{2} $$

you have the answer