$a_n=(1+\frac{1}{2})(1+\frac{1}{4})\dots (1+\frac{1}{2^n}) $ converge Hi guys I have a problem that i need to prove that the sequence:
$$a_n=(1+\frac{1}{2})(1+\frac{1}{4})\dots (1+\frac{1}{2^n}) $$ converges
I need to show its a monotone sequence with an upper and lower bound.
For lower bound, 0 seems to fit nicely.
I can also see that I always multiply by something bigger than one so its always increasing.
My only issue was finding an upper bound, with my calculator I figured that something around e is the upper bound but I can't prove it.
I attempted by induction:
$$ \left(1+\frac{1}{2}\right)\cdot\left(1+\frac{1}{4}\right)\dots\left(1+\frac{1}{2^{n-1}}\right)\left(1+\frac{1}{2^{n}}\right)<4 $$
after multiplying both sides I got :
$$ \left(1+\frac{1}{2}\right)\cdot\left(1+\frac{1}{4}\right)\dots\left(1+\frac{1}{2^{n-1}}\right)\left(1+\frac{1}{2^{n}}\right)\left(1+\frac{1}{2^{n+1}}\right)<4\left(1+\frac{1}{2^{n+1}}\right)<4\cdot2=8 $$
But it seems wrong and Im kind of lost.
Also tried using  $$log_2a_n=log(3)-log(2)+log(5)-log(4)+log(9)-log(8)+\dots log(2^n+1)-log(2^n)$$
No matter how much I try, I just cant prove an upper bound, any suggestions?

Note
\begin{eqnarray}
a_n&=&(1+\frac{1}{2})(1+\frac{1}{4})\dots (1+\frac{1}{2^n})\\
&=&\frac{(1-\frac12)(1+\frac{1}{2})(1+\frac{1}{4})\dots (1+\frac{1}{2^n})}{1-\frac12}\\
&=&2(1-\frac{1}{2^2})(1+\frac{1}{2^2})\dots (1+\frac{1}{2^n})\\
&=&\cdots\\
&=&2(1-\frac{1}{2^{2n}})
\end{eqnarray}
and hence $\{a_n\}$ converges.
