While reading an article on percolation I found this equality. The authors simply said to observe this equality, as it was somewhat obvious. How can one show this?
Show that $\lim_{k \to \infty}\left(1-\prod_{i=3}^k\left(1-\frac{1}{i\ln i}\right)\right)^{\sqrt{\ln k}} = \lim_{k \to \infty} e^{-\frac{1}{\sqrt{ln k}}}$
I can say that, as $\ln(1-x) \leq -x$, we have that $$\sum_{i=3}^\infty \ln\left(1-\frac{1}{i\ln i}\right)\leq \sum_{i=3}^\infty -\frac{1}{i \ln i} = -\infty$$ where we see that the series diverges by the integral test for convergence.
and so I would have that $\lim_{k \to \infty}\prod_{i=3}^k\left(1-\frac{1}{i \ln i}\right) = 0$, BUT there's that $\sqrt{\ln k}$ which complicates things. And besides that, it doesn't suffice for me to show that both limits are equal to 1, I need to understand how we got to $e^{-\frac{1}{\sqrt{\ln k}}}$.
After some work I found this: It is true that $\lim_n\left(1+\frac{a_n}{n}\right)^n = e^{\lim_n a_n } = \lim_n e^{a_n}$ and so: $$\lim_{k \to \infty}\left(1-\prod_{i=3}^k\left(1-\frac{1}{i\ln i}\right)\right)^{\sqrt{\ln k}} =\lim_{k \to \infty}\left[\left(1-\frac{k}{k}\prod_{i=3}^k\left(1-\frac{1}{i\ln i}\right)\right)^k\right]^{\frac{\sqrt{\ln k}}{k}} =$$ $$=\lim_{k \to \infty} \exp\left[-k\prod_{i=3}^k \left(1-\frac{1}{i \ln (i)}\right)\right]^{\frac{\sqrt{\ln{k}}}{k}} = \lim_{k \to \infty} \exp\left[-\sqrt{\ln(k)}\prod_{i=3}^k \left(1-\frac{1}{i \ln(i)}\right)\right]$$
So I only need to show that $\sqrt{\ln(k)}\prod_{i=3}^k\left(1-\frac{1}{i \ln(i)}\right)\approx \frac{1}{\sqrt{\ln(k)}}$
How can one find that $\prod_{i=3}^k\left(1-\frac{1}{i \ln(i)}\right)\approx \frac{1}{\ln(k)}$?
Thanks in advance.