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..
 A: 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.
A: I will demonstrate why the statement doesn't hold for $\alpha = 1$. If you understand my proof, it will be trivial to see that the statement doesn't hold for $\alpha <1$ either.
$\lim_{n \to \infty} X_n = 0$ is in this case the same as saying that for some $N, n>N$ implies $X_n = 0$, because the only alternative is that $X_n=n $ i.o. in which case $X_n$ doesn't converge.
Therefore, if we can show that the probability of such an $N$ existing is not one, then $X_n$ does not converge a.s.Therefore we will calculate the probability for a given $N$, that $X_n=0$ for all $n>N$.
$\mathbb{P}\big(\bigcap_{n>N} X_n = 0\big)=
\prod_{n>N}\mathbb{P}\big(X_n = 0\big)=
\prod _{n>N}1-\frac {1}{n}$
now, we want to use the taylor expansion of $\ln(x)$ at $x=1$ to understand this product.
$$\ln(x) = (x-1)-\frac 12(x-1)^2+...$$
using big O notation, all of the terms can be pulled together so we have
$$\ln(x)=(x-1)+O((x-1)^2)$$
therefore the natural log of our earlier product can be given in the following form:
$$\ln \prod _{n>N}1-\frac 1n = 
\sum_{n>N} \ln(1-\frac 1n)=
\sum_{n>N}-\frac 1n+O(\frac 1{n^2})=
\sum_{n>N}-\frac 1n+\sum_{n>N}O(\frac {1}{n^2})$$
now, remembering that the big O in our representation of the natural log used only a single constant bound, $\sum_{n>N}O(\frac {1}{n^2})$ can be interchanged to $O(\sum_{n>N}\frac {1}{n^2})$ which is a constant, while $$\sum_{n>N}-\frac{1}{n}$$ diverges to negative infinity like
the negative natural log.
Since this sum, which is the natural log of the probability we seek to evaluate, tends to $-\infty$, then the probability itself must be zero.
For clarity, we will denote the event $X_n=0 \forall n>N$ as $E_n$. Now, for any $N$, we have $\mathbb{P}(E_n)=0$. So the event that there is an $N$ such that $ E_N$ occurs, is a union of events of probability zero, so it has probability zero.
In other words, $X_n$ diverges almost surely, in the case $\alpha = 1$.
