Proving a strict inequality in the limit I want to prove that 
$$ \lim_{k \to \infty} \left( 1 + \frac{1}{2} \right) \left(1 + \frac{1}{4} \right)...\left( 1 + \frac{1}{2^k} \right) < e .$$
Using the $AM-GM$ inequality we arrive at
$$\left( 1 + \frac{1}{2} \right) \left(1 + \frac{1}{4} \right)...\left( 1 + \frac{1}{2^k} \right) < \left(\frac{k + 1 - \frac{1}{2^k} }{k} \right)^k = \left( 1 + \frac{1}{k} - \frac{1}{k2^{k}}\right)^k < \left(1 + \frac{1}{k} \right)^k < e.$$
The first inequality is strict because the terms are different.However, I know that in the limit, strict inequalities can transform into equalities. Since the limit of $\left( 1 + \frac{1}{k} - \frac{1}{k2^{k}}\right)^k$ when $k$ goes to infinity is also $e$, how could I prove a strict inequality?

$$ \sum_{k\geq 1}\log\left(1+\frac{1}{2^k}\right)<\sum_{k\geq 1}\frac{1}{2^k}=1,$$
then exponentiate both sides. There is also a simple strengthening, namely
$$ \sum_{k\geq 1}\log\left(1+\frac{1}{2^k}\right)<\sum_{k\geq 1}\left(\frac{1}{2^k}-\frac{1-\log 2}{4^k}\right)=\frac{2+\log 2}{3},$$
leading to
$$ \prod_{k\geq 1}\left(1+\frac{1}{2^k}\right) < 2^{1/3}e^{2/3}.$$
$ \prod_{k\geq 1}\left(1+\frac{1}{2^k}\right) <\frac{31}{13}$ is a more challenging inequality.
