$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?
 A: In general infinite products are studied via the associated series of logarithms. In other words
$$
a_n = \prod_{i=1}^n \left( 1+ \frac{1}{2^i} \right) = \prod_{i=1}^n e^{\log\left( 1+ \frac{1}{2^i} \right)} = e^{\sum_{i=1}^n \log\left(1+\frac{1}{2^n}\right)},
$$
and $\sum \log(1+\frac{1}{2^n})$ converges if and only of $\sum \frac{1}{2^n}$ converges by the limit comparison test.
More generally, if $x_n\geq 0$ then
$$
\prod_{n=0}^\infty\left(1+x_n\right)\text{ converges if and only if } \sum_{n=0}^\infty x_n \text{converges.}
$$
I highly recommend reading the wikipedia article on infinite products.
A: 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.
A: For the lower bound :
Using Bernoulli's inequality we have $x\geq 0$:
$$(1+x)^{\frac{1}{2}}\leq 1+\frac{1}{2}x$$
So :
$$(1+\frac{1}{2})^{\frac{1}{2}}(1+\frac{1}{4})^{\frac{1}{2}}\cdots(1+\frac{1}{2^n})^{\frac{1}{2}}< (1+\frac{1}{4})(1+\frac{1}{8})\cdots(1+\frac{1}{2^{n+1}})$$
Now we can conclude easily
