Convergence with probability one of $\sum \frac{1}{n}X_n$ and $\sum \frac{1}{\sqrt n}X_n$ if $X_n$ are i.i.d. $N(0,1)$ Suppose $X_n$ are $N(0,1)$ i.i.d.:

*

*If $Y_n = \frac{1}{n}X_n$ then does $\sum Y_n$ converge with P.1?


*If $Y_n = \frac{1}{\sqrt n}X_n$ then does $\sum Y_n$ converge with P.1?


*If we have convergence, then prove the limiting distribution is infinitely divisible.
Now, (1) seems like easy task for 2 series Kolmogorov's theorem:

*

*$\sum \mathbb{E} Y_n = 0$

*$\sum \mathbb{Var} Y_n = \sum \frac{1}{n^2}\mathbb{Var} X_n < \infty$
so (1) does converge. I think the limiting distribution is $N(0,\frac{1}{n^2})$ and it is infinitely divisible because $N(0, \frac{1}{n^2}) \stackrel{D}{=} N(0,\frac{1}{n^3}) + \dots + N(0,\frac{1}{n^3})$ n times. I don't know if that is correct though.
I am stuck on (2) too. I tried 3 series theorem with $c = 1$ so:
$$
\sum \mathbb{P}(|Y_n| > 1) = \sum \mathbb{P}(|X_n| > n^2) = \sum \mathbb{P}(X_n > n^2) + \mathbb{P}(X_n < -n^2) = \sum 2\mathbb{P}(X_n > n^2) = \sum 2(1 - \phi(n^2))
$$
Here I got stuck. Of course $2(1 - \phi(n^2)) \to 0$ as $n \to \infty$ but I am not sure about the convergence of such series. Could you give me a hand?
 A: $\sum\limits_{n=1}^{N} \frac 1 {\sqrt n} X_n$ has normal distribution with mean $0$ and variance $\sum\limits_{n=1}^{N} \frac 1 n$. From this it is easy to see that $\sum\limits_{n=1}^{N} \frac 1 {\sqrt n} X_n$ does not even converge in distribution.
Linear combinations of jointly normal random variables have normal distribution  and limits in distribution of infinitely divisible distributions are infinitely divisible distributions. So if we have convergence the the limiting distribution is infinitely divisible .
[ Let $Z_n \sim N(0,r_n)$ with $r_n \to \infty$. Then $\frac {Z_n} {\sqrt {r_n}} \sim N(0,1)$ so $\frac {Z_n} {\sqrt {r_n}} $ converges in distribution. Since $Z_n= \sqrt {r_n} \frac {Z_n} {\sqrt {r_n}} $ it should be clear that $Z_n$ cannot converge in distribution. I will leave the details to you].
A: For $Y_n = \frac{X_n}{\sqrt{n}}$, to check for convergence in probability of $S_n$ (since $E Y_n = 0, Var Y_n = \frac{1}{n})$ and using CLT (with $n E Y_1 = 0, \sqrt{n Var Y_1} = 1$)
$$
P(|S_n-0|>\varepsilon)  = P(S_n >\varepsilon) + P(S_n < -\varepsilon) = 1- (\Phi(\varepsilon) - \Phi(- \varepsilon))> 0
$$
the last inequality is due to the fact that $\varepsilon>0$ and symmetry of $Z \sim N(0,1)$ around its mean. So the limiting probability doesn't converge to 0, hence $S_n \not\to_p 0$. Therefore, it doesn't converge a.s. either.
