Weak LLN Practice

My intended answer to a weak LLN problem on Math.SE. Problem: Suppose $(X_n)$ is a sequence of r.v’s satisfying $P(X_n=\pm\ln (n))=\frac{1}{2}$ for each $n=1,2\dots$. I am trying to show that $(X_n)$ satisfies the weak LLN. The idea is to show that $P(\overline{X_n}>\varepsilon)$ tends to 0, but I am unsure how to do so. My solution: As in the accepted answer in OP’s previous question https://math.stackexchange.com/q/3021650/290189, I’ll assume the independence of $(X_n)$. [Read More]

Solution to a $p$-test Exercise

I intended to answer Maddle’s $p$-test question, but T. Bongers has beaten me by two minutes, so I posted my answer here to save my work. The problem statement This is the sum: $$\sum\limits_{n=3}^\infty\frac{1}{n\cdot\ln(n)\cdot\ln(\ln(n))^p}$$ How do I tell which values of $p$ allow this to converge? The ratio test isn’t working out for me at all. Unpublished solution The integral test will do. $$\begin{aligned} & \int_3^{+\infty} \frac{1}{x\cdot\ln(x)\cdot\ln(\ln(x))^p} \,dx \\ &= \int_3^{+\infty} \frac{1}{\ln(x)\cdot\ln(\ln(x))^p} \,d(\ln x) \\ &= \int_3^{+\infty} \frac{1}{\ln(\ln(x))^p} \,d(\ln(\ln(x))) \\ &= \begin{cases} [\ln(\ln(\ln(x)))]_3^{+\infty} & \text{if } p = 1 \\ \left[\dfrac{[\ln(\ln(x))}{p+1}]^{p+1} \right]_3^{+\infty} & \text{if } p \ne 1 \end{cases} \end{aligned}$$ When $p \ge 1$, the improper integral diverges. [Read More]