Trying to figure out why for $n>1$ it's true that $\prod_{j=n}^{\infty} (1-2^{-j+1}) \geq 1/4$. I'm trying to figure out why for $n>1$ it's true that $\prod_{j=n}^{\infty} (1-2^{-j+1}) \geq 1/4$.
Any hints/answers/tips are greatly appreciated.
 A: It is sufficient to prove the inequality for $n=2$. Using the mathematical induction one can prove
$$(1+x_1)(1+x_2) \ldots (1+x_k) \ge 1 + x_1 + x_2 + \ldots +x_k,$$
where $x_k$ are of the same sign and $x_k > -1$, $k \in \mathbb{N}$. Therefore
$$(1-2^{-2})(1-2^{-3}) \ldots (1-2^{-m}) \ge 1 - 2^{-2} - \ldots - 2^{-m} > \frac{1}{2},$$
i.e.
$$(1-2^{-1})(1-2^{-2})(1-2^{-3}) \ldots (1-2^{-m}) > \frac{1}{4}.$$
Since this infinite product converges
$$\prod_{n=1}^{\infty} (1-2^{-n}) \ge \frac{1}{4}.$$
A: The induction assumption is that the product of the terms from $\left(1-\frac{1}{2}\right)$ up to $\left(1-\frac{1}{2^{m-1}}\right)$ is $\ge \frac{1}{4}\left(1+\frac{1}{2^{m-2}}\right)$. This is true at $m=2$.
For the induction step, suppose the inequality holds at $k$. We show it holds at $k+1$. So we need to show that
$$\frac{1}{4}\left(1+\frac{1}{2^{k-2}}\right)\left(1-\frac{1}{2^k}\right)\ge \frac{1}{4}\left(1+\frac{1}{2^{k-1}}\right).$$
This is not hard to verify. Get rid of the irrelevant $1/4$ part, and expand. We need to show that
$$\frac{1}{2^{k-2}}-\frac{1}{2^k}-\frac
{1}{2^{2k-2}}\ge \frac{1}{2^{k-1}},$$
which is true.  
Remark: We used the trick, common in proving inequalities by induction, of strengthening the induction assumption. The inequality we used is tight at $m=2$ and $m=3$.  
A: For $0\le x,y\le1$, $(1-x)(1-y)=1-x-y+xy\ge1-(x+y)$. Inductively, we can show that for $0\le x_j\le1$, we have
$$
\prod_j(1-x_j)\ge1-\sum_jx_j
$$
Therefore, for $n\ge1$,
$$
\begin{align}
\prod_{j=n}^\infty\left(1-2^{-j+1}\right)
&\ge1-\sum_{j=n}^\infty2^{-j+1}\\
&=1-\frac{2^{-n+1}}{1-2^{-1}}\\[9pt]
&=1-2^{-n+2}
\end{align}
$$
Thus,
$$
\begin{align}
\prod_{j=2}^\infty\left(1-2^{-j+1}\right)
&=\left(1-2^{-1}\right)\prod_{j=3}^\infty\left(1-2^{-j+1}\right)\\
&\ge\frac12\cdot\frac12\\
&=\frac14
\end{align}
$$
A: The line $y=1+\frac12x$ intersects the convex graph $y=e^x$ at $(0,1)$ and a second point $(x_0,y_0)$.
We have $x_0<-1$ because $\frac1e<\frac12$.
Therefore $1+\frac12x>e^x$ for $-1\le x<0$, i.e. 
$$ \ln(1-x)>-2x\qquad\text{for }0<x\le \frac12.$$
Therefore 
$$\ln(1-2^{-j+1}) >-2^{2-j}\qquad\text{for }j\ge2$$
and so
$$\ln\prod_{j=n}^\infty(1-2^{-j+1})\ge -\sum_{j=n}^\infty 2^{2-j}=-2^{3-n}$$
shows the convergence of the product and the estimate
$$ \prod_{j=n}^\infty(1-2^{-j+1})\ge e^{-2^{3-n}}\ge 1-2^{3-n}$$
(using $e^x\ge 1+x$ on the right).
Then for $n\ge 2$ we have
$$\begin{align} \prod_{j=n}^\infty(1-2^{-j+1})&\ge \prod_{j=2}^\infty(1-2^{-j+1})\\&=\left(1-\frac12\right)\left(1-\frac14\right)\left(1-\frac18\right)\left(1-\frac{1}{16}\right)\prod_{j=6}^\infty(1-2^{-j+1})\\&\ge \frac12\cdot\frac34\cdot\frac78\cdot\frac{15}{16}\cdot\frac78\\&=\frac {2205}{8192}>\frac14.\end{align}$$
