If an infinite product$\ P= 0$ times a constant and an unbounded function gives$\ 1$, will the same for$\ Q< \infty$ certainly be$\ >1$?

Question

Let$\ P=\lim_{n\to \infty} \prod_{i=1}^n x_i=0$, $\ Q=\lim_{n\to \infty} \prod_{i=1}^n y_i< \infty$ and$\ \lim_{n\to \infty} f(n)= \infty$. If, for some real$\ k$, $\ \lim_{n \to \infty} k f(n) \prod_{i=1}^n x_i=1$ and$\ y_i>x_i>0$, will $\ \lim_{n \to \infty} k f(n) \prod_{i=1}^n y_i>1$ be certainly true? If not, in which case(s) it isn't?

Answer 1

Yes, and the hypothesis $\lim_{n\to \infty} \prod_{i=1}^n y_i< \infty$ isn't needed.

The hard part is proving that $\lim_{n \to \infty} k f(n) \prod_{i=1}^n y_i$ exists at all, rather than oscillating or something like that. Write $$ \lim_{n \to \infty} k f(n) \prod_{i=1}^n y_i = \lim_{n \to \infty} \bigg( \Big( k f(n) \prod_{i=1}^n x_i \Big) \Big( \prod_{i=1}^n \frac{y_i}{x_i} \Big) \bigg). $$ The first factor on the right-hand side converges to $1$ by assumption. Since each $\frac{y_i}{x_i}>1$, the second factor is strictly increasing as $n$ increases, hence either converges or diverges to $+\infty$. Therefore the limit in question also either converges or diverges to $+\infty$.

Since $k f(n) \prod_{i=1}^n y_i = k f(n) \frac{y_1}{x_1} \cdot x_1 \prod_{i=2}^n y_i \ge k f(n) \frac{y_1}{x_1} \prod_{i=1}^n x_i$, we see that $\lim_{n \to \infty} k f(n) \prod_{i=1}^n y_i$ must be at least $\frac{y_1}{x_1}$, which is strictly greater than $1$.