Does the series $\sum\limits_{n\geq3}P\{\xi_n>\sqrt{2\ln n+2\ln\ln n}\}$ converges?

Question

Let $\xi_n$ be a sequence of the random variables with standard normal distribution. Does the following series $$\sum\limits_{n\geq3}P\{\xi_n>\sqrt{2\ln n+2\ln\ln n}\}$$ converges?

Answer 1

If $A>0$ then $$\int_A^\infty e^{-t^2/2}\le A^{-1}\int_A^\infty te^{-t^2/2}=A^{-1}e^{-A^2/2}.$$Hence $$P\{\xi_n>\sqrt{2\ln n+2\ln\ln n}\} \le\frac c{\sqrt{\ln n}}e^{-(\ln n+\ln\ln n)}=\frac c{n(\ln n)^{3/2}},$$so the sum converges.

Bonus: At first blush that upper bound on the integral look fairly rough. But it's actually not far from the truth, since the exponential dies so fast. If $A>1$ then $(A+1/A)^2<A^2+4$, so $$\int_A^\infty e^{-t^2/2}\ge\int_A^{A+1/A}e^{-t^2/2}\ge A^{-1}e^{-(A+1/A)^2/2}\ge\frac 1{e^2A}e^{-A^2/2}.$$

Which means we can pin this one down:

Exercise: $\sum P\{\xi_n>\sqrt{2\ln n+\alpha\ln\ln n}\}$ converges for $\alpha>1$ and diverges for $\alpha\le1$.