how to prove that this series bounded and monotonic..

**transgalactic** · December 10th 2008, 12:12 PM

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?

**Mathstud28** · December 10th 2008, 02:18 PM

Originally Posted by transgalactic

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 $\prod_{k=1}^{n}\bigg[1-\frac{1}{2^k}\bigg]$ . Boundedness is simple

$\forall{n}\in(1,\infty)~1-\frac{1}{2^k}<1\implies\prod_{k=1}^{n}\bigg[1-\frac{1}{2^k}\bigg]<\prod_{k=1}^{n}1=1$

As for monotonicity you are making a mistake we are talking about the monotonicty of $a_n$ , that is $a_n=\prod_{k=1}^{n}b_k$ NOT $\prod_{k=1}^{n}a_k$ . So we need to show that $\prod_{k=1}^{n+1}\bigg[1-\frac{1}{2^k}\bigg]<\prod_{k=1}^{n}\bigg[1-\frac{1}{2^{k}}\bigg]$ . Since both sides are positive for all n we have that

$\prod_{k=1}^{n+1}\bigg[1-\frac{1}{2^k}\bigg]<\prod_{k=1}^{n}\bigg[1-\frac{1}{2^{k}}\bigg]\implies\displaystyle{\frac{\prod_{k=1}^{n+1}\bigg[1-\frac{1}{2^k}\bigg]}{\prod_{k=1}^{n+1}\bigg[1-\frac{1}{2^k}\bigg]}<\frac{\prod_{k=1}^{n}\bigg[1-\frac{1}{2^k}\bigg]}{\prod_{k=1}^{n+1}\bigg[1-\frac{1}{2^k}\bigg]}}$ . Simplifying gives $1<\frac{1}{1-\frac{1}{2^{n+1}}}$ which is obviously true for all positive n.

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