Let $a > 0$ and $\{y_n\}_{n\geq0}$ be a sequence such that $y_0 > 0$, $y_n > a \ \forall \ n > 1$ and
$$\sum_{n=1}^\infty \frac 1 {y_n} \to \infty$$
Prove That
$$ \lim_{n\to \infty} \prod_{k=1}^n \left( 1 - \frac a {y_k}\right) = 0 $$
I tried to expand the product using Vieta's formula, but I'm getting the product to be infinite.
Edit : I'm adding the source of the problem. I found it at the end of Theorem 1 on page 7 of this PDF.
Here is a screenshot
Any help will be appreciated.
By the strict convexity of the exponential function, we have
$$1 - x \leqslant e^{-x}\tag{1}$$
for all $x\in \mathbb{R}$, and the inequality is strict for all $x \neq 0$. The assumptions yield $0 < 1 - \frac{a}{y_k} \leqslant 1$ for $k > 1$ (probably a typo and it should have been $\geqslant 1$, but that doesn't matter). Using $(1)$, we find
$$0 < \prod_{k = 2}^m \biggl(1 - \frac{a}{y_k}\biggr) \leqslant \exp \Biggl(-a \sum_{k = 2}^m \frac{1}{y_k}\Biggr)\tag{2}$$
for all $m \geqslant 2$. The assumption $\sum \frac{1}{y_k} = +\infty$ implies that the right hand side of $(2)$ tends to $0$ as $m \to \infty$, hence by squeezing we deduce
$$\lim_{m\to \infty} \prod_{k = 2}^m \biggl(1 - \frac{a}{y_k}\biggr) = 0.$$
Multiplying with the constant $1 - \frac{a}{y_1}$ yields
$$\lim_{m\to\infty} \prod_{k = 1}^m \biggl(1 - \frac{a}{y_k}\biggr) = 0.$$