Does limit of $x_n=(1+\frac{1}{2})*(1+\frac{1}{4})*...*(1+\frac{1}{2^n})$ exist? I tried to show that the sequence is increasing and limited, but couldn't find the limit.
I also tried with squeeze theorem, but $(1+\frac{1}{2^n})^n<=x_n<=(1+\frac{1}{2})^n$ is not helping and I ran out of ideas.
 A: A handy fact about products like this is
$$ 1+\sum_{k=1}^n a_k \le \prod_{k=1}^n (1+a_k) \le \exp\Big(\sum_{k=1}^n a_k\Big) $$
(The first inequality is a generalization of Bernoulli's inequality; the second is $1+x\le e^x$ used $n$ times.)  So you can check the convergence of this kind of product by checking the convergence of a related series, for which we have a bunch of standard techniques.  In this case, the related series is a familiar one.
A: $$A_N=\prod_{n=1}^{N}\left(1+\frac{1}{2^n}\right)=\exp\sum_{n=1}^{N}\log\left(1+\frac{1}{2^n}\right)\leq\exp\sum_{n=1}^{N}\frac{1}{2^n}\leq e.$$
Since the LHS is increasing and bounded, the limit
$$\lim_{N\to +\infty} A_N = \prod_{n=1}^{+\infty}\left(1+\frac{1}{2^n}\right)$$
exists.
A: I'll write it this way:
Let : $ S_n = ln(x_n) = \sum_{k=0}^n ln(1 + \frac{1}{2^k}) $
As I said: $ a_n = ln(1+\frac{1}{2^n}) \leq \frac{1}{2^n} = v_n $
You get : $ S_n \leq \sum_{k=1}^n v_k \leq \sum_{k=1}^{+\infty} v_k = 1$
$(S_n)$ is increasing, bounded so it converges.
