Let $X_n$ be a sequence of r.v.s satisfying \begin{equation*} \lim_{n\rightarrow\infty}E[X_n]=0\qquad \lim_{n\rightarrow\infty}\operatorname{Var}[X_n]=0. \end{equation*} Can we prove $X_n\rightarrow0$ almost surely? Or what additional conditions can we add to make it right?
My original question comes from two exercises about SDE:
1) Let $X_t=\sigma e^{-\lambda t}\int_0^te^{\lambda s}dB_s$ be a solution of Ornstein-Uhlenbeck equation, where $\sigma$ and $\lambda$ are both constants, then we want to compute \begin{equation*} \limsup_{t\rightarrow\infty}\frac{X_t}{\sqrt{\ln{t}}}\quad\text{ and }\quad\liminf_{t\rightarrow\infty}\frac{X_t}{\sqrt{\ln{t}}}. \end{equation*}
2) Let $Y_t=(1-t)\int_0^t\frac{dB_s}{1-s}$ be the Brownian bridge, then we want to show $Y_t\rightarrow 0$ almost surely as $t\rightarrow 1$.
It is not hard to verify that $\frac{X_t}{\sqrt{\ln{t}}}$ and $Y_t$ both have 0 expectations and variance tends to 0, so what's the difference between them and how to work them out?
