<s>[INST] 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?
 [/INST] You can just ignore the first term and carry on with your method from $i=2$.
Then you get $\prod\limits_{i=2}^ka_i\le\cdots\le\left(1+\dfrac 1{2(k-1)}\right)^{k-1}\le e^{0.5}$
And then $\prod\limits_{i=1}^\infty a_i=\left(1+\dfrac 12\right)\prod\limits_{i=2}^\infty a_i\le 1.5\,e^{0.5}<e$
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