$a_n=(1+\frac{1}{2})(1+\frac{1}{4})\dots (1+\frac{1}{2^n}) $ converge

Question

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?

Answer 1

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.

Answer 2

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

Answer 3

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.