Xn=(1-1/2)(1-1/4)..(1-(1/(2^n))
i tried to prove that its monotonic
by :
1-1/(2^n) = (2^n-1)/2^n
2^n -1 <2^n
obviously its correct
the numerator of each object is smaller then the denominator.
what now??
and how to prove that its bounded?
Xn=(1-1/2)(1-1/4)..(1-(1/(2^n))
i tried to prove that its monotonic
by :
1-1/(2^n) = (2^n-1)/2^n
2^n -1 <2^n
obviously its correct
the numerator of each object is smaller then the denominator.
what now??
and how to prove that its bounded?
Is there a reason you keep calling these series instead of products?
You want to prove monotonicity and boundedness of $\displaystyle \prod_{k=1}^{n}\bigg[1-\frac{1}{2^k}\bigg]$. Boundedness is simple
$\displaystyle \forall{n}\in(1,\infty)~1-\frac{1}{2^k}<1\implies\prod_{k=1}^{n}\bigg[1-\frac{1}{2^k}\bigg]<\prod_{k=1}^{n}1=1$
As for monotonicity you are making a mistake we are talking about the monotonicty of $\displaystyle a_n$, that is $\displaystyle a_n=\prod_{k=1}^{n}b_k$ NOT $\displaystyle \prod_{k=1}^{n}a_k$. So we need to show that $\displaystyle \prod_{k=1}^{n+1}\bigg[1-\frac{1}{2^k}\bigg]<\prod_{k=1}^{n}\bigg[1-\frac{1}{2^{k}}\bigg]$. Since both sides are positive for all n we have that
$\displaystyle \prod_{k=1}^{n+1}\bigg[1-\frac{1}{2^k}\bigg]<\prod_{k=1}^{n}\bigg[1-\frac{1}{2^{k}}\bigg]\implies\displaystyle{\frac{\prod_{k=1}^{n+1}\bigg[1-\frac{1}{2^k}\bigg]}{\prod_{k=1}^{n+1}\bigg[1-\frac{1}{2^k}\bigg]}<\frac{\prod_{k=1}^{n}\bigg[1-\frac{1}{2^k}\bigg]}{\prod_{k=1}^{n+1}\bigg[1-\frac{1}{2^k}\bigg]}}$. Simplifying gives $\displaystyle 1<\frac{1}{1-\frac{1}{2^{n+1}}}$ which is obviously true for all positive n.