Prove that a sequence converges but not to $0$ 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 :).
 A: Note that $\log(1-x) \ge -x -{1 \over 2} x^2$ for $x \in [0,1)$.
$\log x_{n+1} = \log x_n + \log (1-{1 \over 2^{n+1}}) \ge \log x_n  - ({1 \over 2^{n+1}} + {1 \over 2} {1 \over 2^{2n+2}})$.
Since the rightmost terms are summable, it follows that
$\log x_n \ge -B$ for some $B>0$ and hence $x_n \ge {1 \over e^B}$ for all $n$.
A: An approach that doesn't require taking a logarithm like in copper.hat's nice answer: Notice that
$$x_n - x_{n + 1} = \frac{x_n}{2^{n + 1}}$$
Given the fact that $x_n > 2^{-n}$ (as can be seen by noting that almost all terms in the product are larger than $1/2$), we can iterate this identity,
$$x_0 - x_1 \ge \frac{1}{2^1}$$
$$x_1 - x_2 \ge \frac{1}{2^3}$$
$$x_2 - x_3 \ge \frac{1}{2^6}$$
and in general*,
$$x_k - x_{k + 1} \ge \frac{1}{2^{(k + 1)(k + 2) / 2}}$$
Combining these, we find that
$$x_0 - x_k \ge \frac 1 2 + \frac 1 {2^3} + \frac 1 {2^6} + \cdots + \frac 1 {2^{k(k + 1) / 2}}$$
By comparison to a geometric series, this difference is always less than $1 - 1/2^2 = 3/4$; hence, $x_n > \frac 1 4$ for every $n$.

*The exponents are the $(k + 1)$-th triangular numbers. Convince yourself why.
A: 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 $2k+2\ge k+3$.
A: $$ x-1 \ge \log_e x \qquad \text{and “}{=}\text{'' only if } x=1. $$
That's how that inequality is usually stated, but with the substitution $u=x-1$ (so that $x=u+1$), it becomes $$ u \ge \log_e(u+1) \qquad \text{and “}{=}\text{'' only if } u=0. $$
Then we have
\begin{align}
& \log_e \Big( (1 + 2^{-1})(1 + 2^{-2})(1 + 2^{-3}) \cdots \Big) \\[10pt]
= {} & \log_e(1+2^{-1}) + \log_e(1 + 2^{-2}) + \log_e (1 + 2^{-3}) + \cdots \\[10pt]
\ge {} & -\Big( 2^{-1} + 2^{-2} + 2^{-3} + \cdots \Big) = -1 > -\infty.
\end{align}
Can you show that if the infinite product you started with converged to $0$ then the logarithm above would be $-\infty$ rather than ${}>-\infty$?
