Does the following sequence converges a.s.? This is an example from Gut: A probability course.
Let $\alpha>0, X_1, X_2,..$be independent r.v.s. such that $$P(X_n=0)=1-1/n^\alpha$$ and $$P(X_n=n)=1/n^\alpha$$, $n\geq 1$
Gut claims that $X_n\rightarrow 0$ a.s. iff $$\alpha>1$$. But it looks to me that even for $0<\alpha\leq 1$ we have a.s. convergence. What is wrong with my observation? Thank you..

This is a direct application of Borel-Cantelli lemmas. 
$\displaystyle \sum_{n=1}^{\infty} P(X_n = n) = \displaystyle \sum_{n=1}^{\infty} \frac{1}{n^{\alpha}} <\infty$ iff $\alpha > 1$. Therefore, by Borel-Cantelli lemma $P(X_n = n \text{ i.o.} )= 0$ (i.o means infinitely often). Hence, $X_n$ converges to 0 a.s.
$\displaystyle \sum_{n=1}^{\infty} P(X_n = n) = \displaystyle \sum_{n=1}^{\infty} \frac{1}{n^{\alpha}} =\infty$ if $0< \alpha \leq 1$. As we know that $X_i$'s are independent, by Borel-Cantelli lemma $P(X_n = n \text{ i.o.} )= 1$. Hence, $X_n$ does not converge to 0 a.s.
