<s>[INST] Showing that $\displaystyle\prod_{k = 0}^{\infty} \frac{1+kx}{2+kx} = 0$ for any fixed $x \in \mathbb{R}_{>0}$ How can I show that
$$\displaystyle\prod_{k = 0}^{\infty} \frac{1+kx}{2+kx} = 0$$ when $x$ is a positive real number ? 
 [/INST] $$ \prod_{k = 0}^{\infty} \frac{1+kx}{2+kx} = 0$$
Proof:
$$\frac{1+kx}{2+kx}=1+\frac{-1}{2+kx} \leqslant\exp \left( \frac{-1}{2+kx} \right) \leqslant \exp \left( \frac{-x^{-1}}{k} \right)$$
So, 
$$ \prod_{k = 1}^{N} \frac{1+kx}{2+kx} \leqslant \prod_{k = 1}^{N} \exp \left( \frac{-x^{-1}}{k} \right) = \exp(-x^{-1}H_N)$$
where $H_N$ is the $N^{th}$ harmonic number.
Using the simple estimate $H_N > \log N$, we have that
$$\prod_{k = 0}^{N} \frac{1+kx}{2+kx} \le \frac{1}{2} \exp(-x^{-1}\log N)=\frac{1}{2N^{x^{-1}}}\to 0$$
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