We have a graph with n vertices. Each edge can appear with probability $\frac{1}{n\ln(n)}$. Prove that $\lim\limits_{n\rightarrow\infty}Pr[$there is no isolated vertices$] = 0$.
I think that first of all we should connect vecrtices so there won't be isolated vertices. We connect them by pairs $\binom{n}{2}\frac{1}{n\ln(n)} + \binom{n-2}{2}\frac{1}{n\ln(n)} + \cdots + \binom{2}{2}\frac{1}{n\ln(n)}$ and divide it on $\frac{n}{2}$ (because connection does not depends on order). And do smth with other edges $\sum_{i = 1}^{\frac{n(n-1)}{2} - \frac{n}{2}}(\frac{1}{n\ln(n)})^i(1-\frac{1}{n\ln(n)})^{\frac{n(n-1)}{2} - \frac{n}{2}}$.
So the probability is $\frac{(\sum_{i = 1}^{\frac{n(n-1)}{2} - \frac{n}{2}}(\frac{1}{n\ln(n)})^i(1-\frac{1}{n\ln(n)})^{\frac{n(n-1)}{2} - \frac{n}{2} - i})\cdot\frac{(\binom{n}{2}\frac{1}{n\ln(n)} + \binom{n-2}{2}\frac{1}{n\ln(n)} + \cdots + \binom{2}{2}\frac{1}{n\ln(n)})}{\frac{n}{2}}}{\sum_{i = 1}^{\frac{n(n-1)}{2}}(\frac{1}{n\ln(n)})^i(1-\frac{1}{n\ln(n)})^{\frac{n(n-1)}{2} - i}}$. I think that it is alright but I am not sure.
I assume that the edge choices are independent. Put $p=\frac{1}{n\ln(n)}$. A probability that a given vertex is isolated is $$(1-p)^{n-1}\ge 1-(n-1)p\ge 1-\tfrac{1}{\ln n}$$ by Bernoulli's inequality, and the last value tends to $1$.