Comparison result between infinite products

Question

Suppose I have two sequences of positive numbers $(x_n)_{n\in\mathbb{N}}$ and $(y_n)_{n\in\mathbb{N}}$ such that

1. There exists a $\gamma>1$ such that $\gamma^n|x_n-y_n|\rightarrow0$ as $n\rightarrow\infty$.
2. $\prod_{n=1}^{\infty}x_n = 0$.

Question: Can we conclude that $$ \prod_{n=1}^{\infty}y_n = 0? $$

I know that without condition 1. this is definitely untrue. Does the situation change when we have exponential fast convergence?

Answer 1

We can conclude that $\prod\limits_{n = 1}^{\infty} y_n$ diverges to $0$ if $\lim\limits_{n\to\infty} x_n$ exists, or $\liminf\limits_{n\to\infty} x_n > 0$, but not without any further assumptions.

To see that such assumptions are necessary, choose $\lambda > 1$ and let $x_{n} = \lambda^{-n^2},\, y_n = \lambda^{-n}$ when $n$ is a square, $x_n = y_n = \lambda^{k}$ when $k^2 < n < (k+1)^2$. Then $0 \leqslant y_n - x_n < \lambda^{-n}$ for all $n$, so $\gamma^n\lvert x_n - y_n\rvert \to 0$ for all $\gamma \in (1,\lambda)$. We have

$$\prod_{n = k^2}^{k^2+2k} x_n = \lambda^{-k^4 + 2k^2}$$

for all $k$, hence the product of the $x_n$ diverges to $0$. But

$$\prod_{n = k^2}^{k^2+2k} y_n = \lambda^{-k^2+2k^2} = \lambda^{k^2} \geqslant \lambda > 1$$

for all $k$ and

$$\prod_{n = k^2+1}^{(k+1)^2} y_n = \lambda^{2k^2-(k+1)^2} = \lambda^{(k-1)^2-2} \geqslant \lambda^2$$

for $k \geqslant 3$, so the product of the $y_n$ diverges to $+\infty$.

It is clear that we also have $\prod y_n = 0$ if $x_n \to 0$, for then condition 1 forces $y_n \to 0$ too. And if $\liminf x_n > 0$, there is a $\delta \in (0,1]$ and an $n_0 \in \mathbb{N}$ such that $\min \{ x_n, y_n\} \geqslant \delta$ for all $n \geqslant n_0$, whence

$$\lvert \log x_n - \log y_n\rvert \leqslant \delta^{-1}\lvert x_n - y_n\rvert$$

for $n \geqslant n_0$, from which it follows that

$$\sum_{n = 1}^{\infty} \lvert \log x_n - \log y_n\rvert < +\infty$$

and consequently

\begin{align} \prod_{n = 1}^{\infty} x_n = 0 &\iff \sum_{n = 1}^{\infty} \log x_n = -\infty \\ &\iff \sum_{n = 1}^{\infty} \log y_n = -\infty \\ &\iff \prod_{n = 1}^{\infty} y_n = 0\,. \end{align}