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.

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.
