TITLE: Prove that a sequence converges but not to $0$
QUESTION [6 upvotes]: The following is the sequence I came across while solving "Analysis of Euclidean Space" by Kenneth Hoffman.
Let 
$$x_n=(1-2^{-1})(1-2^{-2})\cdots(1-2^{-n})$$
I want to prove that the limit exists and not $0$.
I have already proved that it converges since it is bounded below by $0$ and monotonicly decreasing. But how do I find the limit or prove that the limit is not $0$. Thanks in Advance :).

REPLY [3 votes]: Let $P_n$ be the product of the first $n$ terms. We show by induction that  $P_n\ge\frac{1}{4}+\frac{1}{2^{n+1}}$.
Suppose the result is true for $n=k$. We show it is true for $n=k+1$. We have
$$P_{k+1}\ge \left(\frac{1}{4}+\frac{1}{2^{k+1}}\right)\left(1-\frac{1}{2^{k+1}}\right)=\frac{1}{4}+\frac{1}{2^{k+1}}-\frac{1}{2^{k+3}}-\frac{1}{2^{2k+2}}.$$
To finish, it is enough to show that $\frac{1}{2^{k+3}}+\frac{1}{2^{2k+2}}\le \frac{1}{2^{k+2}}$. This is obvious, since $2k+2\ge k+3$.