Show the sequence $a_1=1$, $a_{n+1}=(1-\frac{1}{2^n})a_n$ converges. $a_1=1 \,\text{and} \,$$a_{n+1}=(1-\frac{1}{2^n})a_n$
My try:This is a decreasing sequence and bounded below by $1$.So $a_n$ converges.
 A: The first elements of the sequence:
$$a_1=1,\,a_2=\frac12,\,a_3=\frac38,\,a_4=\frac{21}{64}\,,\ldots$$
so it looks like a descending sequence. With a little induction
$$a_{n+1}:=\left(1-\frac1{2^n}\right)a_n\le a_n\iff1-\frac1{2^n}\le1$$
and since the last inequality is trivial we're done.
Finally, again with a little induction
$$a_{n+1}=\left(1-\frac1{2^n}\right)a_n\ge0\iff a_n\ge0$$
and thus zero is a lower bound, and thus the sequnece converges.
A: You are correct except the lower bound should not be 1. A lower bound of 0 works though, so by the monotone convergence theorem the sequence converges.
A: Check that,
$$a_n = \prod_{k=1}^{n-1}\frac{a_{k+1}}{a_k} =  \prod_{k=1}^{n-1}(1-\frac{1}{2^k}) $$
Therefore 
$$\lim_{n\to \infty}\ln a_n = \sum_{n=0}^\infty \ln(1-\frac{1}{2^n})$$
which converges since 
$$\lim_{n\to \infty} 2^n \ln(1-\frac{1}{2^n}) = \lim_{h\to 0} \frac{\ln(1-h)}{h} = 1$$
that is 
$$\ln(1-\frac{1}{2^n}) \approx \frac{1}{2^n} $$
$\lim\limits_{n\to \infty}\ln a_n$ exists then  $\lim_{n\to \infty} a_n$ exists
