Showing that $\displaystyle\prod_{k = 0}^{\infty} \frac{1+kx}{2+kx} = 0$ for any fixed $x \in \mathbb{R}_{>0}$

Question

How can I show that $$\displaystyle\prod_{k = 0}^{\infty} \frac{1+kx}{2+kx} = 0$$ when $x$ is a positive real number ?

Answer 1

$$ \prod_{k = 0}^{\infty} \frac{1+kx}{2+kx} = 0$$

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

$$\prod_{k = 0}^{N} \frac{1+kx}{2+kx} \le \frac{1}{2} \exp(-x^{-1}\log N)=\frac{1}{2N^{x^{-1}}}\to 0$$