Prove that the sum of Cauchy distribution random variables divide by a infinite sequence convergence Let $X_1,...,X_n$ are independent identical random variables of standard Cauchy distribution (i.e. the density is $\frac{1}{\pi(1+x^2)}$). Let $n \to \infty$ and consider the random variable $S_n=X_1+,...,+X_n$.

*

*Prove:  $\frac{S_n}{a_n}\rightarrow0$ a.e. iff $\sum \frac{1}{a_n}<\infty$.

*Try to give the necessary and sufficient conditions for which  $$
\frac{\max(X_1,...,X_n)}{a_n}\to 0 \quad\text{a.e.}
$$
My ideas so far:

*

*For problem 1, given  $\sum \frac{1}{a_n}<\infty$, I would like to prove that $\frac{S_n}{a_n}\rightarrow0$: by Kroencker lemma, I just need to prove that $\sum\frac{X_n}{a_n}<\infty.$ Then I do it by applying Kolmogorov three series theorem: thus I can easily prove the convergence of $\sum\frac{X_n}{a_n}$. But how to prove the converse? What can I deduce from $\frac{S_n}{a_n}\rightarrow0$? Maybe I should try to use characteristic function?

*For problem 2, I think maybe there is some way to convert $\frac{\max(X_1,...,X_n)}{a_n}$ to a similar form as in problem 1: am I right?

Thanks in advance for any tips or help in general.
 A: The condition for (1) and (2) is the same, summability of $1/a_n$.
First the easy case: If $\sum 1/a_n$ diverges, then $X_n>a_n$ infinitely often.
Therefore  $R_n:=S_n/a_n$ cannot tends to zero because
$$\frac{X_n}{a_n}= R_n- \frac{a_{n-1}}{a_n}R_{n-1} \, $$
as noted by @p4sch.
Also, under the assumption of divergence of $\sum 1/a_n$, the ratio
$$\frac{\max(X_1,...,X_n)}{a_n} \ge \frac{ X_n}{a_n} $$
does not tend to 0.
Conversely, if $\sum 1/a_n$ converges, then instead of Kronecker's lemma, one can  use a dyadic argument.
Since $Q_\ell:=S_\ell/\ell$ is a standard Cauchy variable for any $\ell \ge 1$,
we have
$$  P\left[\frac{|S_{2^{k+1}}|}{a_{2^k}}>\epsilon\right] = 
P\left[ |Q_{2^{k+1}}| > \epsilon \frac{a_{2^k}}{2^{k+1}} \right]\le C\frac{2^k}{\epsilon a_{2^k}} \quad (*)\,,
$$
where $C$ is a constant.
Now the RHS of (*) is summable in $k$, because for each $k$,
$$ \frac{2^k}{a_{2^k}} \le 2\sum_{j=2^{k-1}+1}^{2^{k}} \frac{1}{a_j} \,.
$$
Thus,
$$\limsup_k \frac{|S_{2^{k+1}}|}{a_{2^k}} \le \epsilon \,, $$
almost surely, so (intersecting these events over $\epsilon=1/m$ for all $m$) we obtain that
$$\limsup_k \frac{|S_{2^{k+1}}|}{a_{2^k}}=0\, $$
almost surely.
To interpolate, use the first Ottaviani-Levy inequality, see the first inequality in A.1.2 (page 431) in [1]. (or see [2], [3]). We obtain
$$   P\left[\max_{n\in (2^k, 2^{k+1}]} \frac{|S_n|}{a_n} \ge \epsilon\right] \le P\left[\max_{n\in (2^k, 2^{k+1}]} |S_n| \ge \epsilon a_{2^k}\right] \le    2 P[  |S_{2^{k+1}}| \ge \epsilon a_{2^k}] 
\le 2C\frac{2^k}{\epsilon a_{2^k}} \,.$$
This is just twice $(*)$, so almost surely,
$$\limsup_n \frac{S_{n}}{a_n}=0\,. $$
Also, in this case,
$$\frac{\max(|X_1|,...,|X_n|)}{a_n}  \to 0
$$
almost surely using the same logic and the second Ottaviani-Levy inequality in the same reference.
[1] https://link.springer.com/content/pdf/bbm%3A978-1-4757-2545-2%2F1.pdf
[2] Billingsley, P. (1995) Probability and Measure, 3rd ed., Theorem 22.5, p. 288.
[3] Kai Lai Chung, A course in probability theory, 3rd ed., P. 126
A: This is not an answer but a remark, that something cannot be true in the Problem, as already pointed out in the comments.
It's an easy exercise to calculate the characteristic function of $\frac{S_n}{a_n}$, denote it by $\varphi_n(t)=e^{-\frac{n|t|}{a_n}}$. Now for instance take the sequence $a_n=n\log n$ (with $a_1=1)$, then we see that $\varphi_n(t)\rightarrow 1$ as $n\rightarrow\infty$, hence the random variables $\frac{S_n}{a_n}$ must converge to $0$, but the series $\sum \frac{1}{a_n}$ is not convergent.
A: First, we should note that the event
$$\{ S_n/a_n \rightarrow 0 \}$$
is an element of the terminal $\sigma$-algebra $\tau((X_n)_{n \in \mathbb{N}})$, since $a_n \rightarrow \infty$. So the probability of this event is $0$ or $1$.
Moreover we can compute $$\mathbb{P}(|S_n|>\delta a_n )$$ explicitly by noting that the the sum of independent standard cauchy distribution has a cauchy distribution with scale parameter $n$. (This can be checked by determining the characteristic function of $S_n$ and Levy's theorem.)
So, using the symmetry of the density, we have
$$\mathbb{P}(|S_n|>\delta a_n ) =\frac{2}{\pi} \int_{\delta a_n}^\infty \frac{n}{n^2+t^2} \, \mathrm{d} t \leq \frac{2}{\pi} \frac{n}{\delta a_n}.$$
Thus, if the series $\sum_{k=1}^\infty n/a_n$ is convergent, we can conclude (by the simple part of the Borel-Cantelli lemma) that $\mathbb{P}\{ S_n/a_n \rightarrow 0 \} =1$ holds.
We see that the our simple argument, based on the Borel-Cantelli lemma, is not strong enough. In fact, we should procced by your argument: $\sum_{n=1}^\infty X_n/a_n$ converges a.e. if
$$\sum_{n=1}^\infty \mathbb{P}(|X_n|> a_n) <\infty.$$
As we have already seen, we have
$$\sum_{n=1}^\infty \mathbb{P}(|X_n|> a_n) \leq = \frac{2}{\pi} \sum_{n=1}^\infty \int_{a_n} \frac{1}{1+t^2} \, \mathrm{d} t \leq \frac{2}{\pi} \sum_{n=1}^\infty \frac{1}{a_n}.$$
So, if $\sum_{n=1}^\infty \frac{1}{a_n} < \infty$, then $\sum_{n=1}^\infty X_n/a_n$ converges a.e. and now Kronecker's lemma implies that $\frac{1}{a_n} \sum_{k=1}^n X_k \rightarrow 0$ a.e., as claimed.
If $\sum_{n=1}^\infty 1/a_n$ is divergent, we note that
$$\mathbb{P}(|X_n| > a_n) = \frac{2}{\pi} \int_{a_n}^\infty \frac{1}{1+t^2} \,\mathrm{d} t = \frac{2}{\pi}(1- \arctan(a_n)).$$
Thus, we find that $\sum_{n=1}^\infty \mathbb{P}(|X_n| > a_n)$ is divergent as well. By the second part of the Borel-Cantelli lemma, we conclude that $$\mathbb{P}(|X_n| > a_n \text{ for infinty many } n \in \mathbb{N}) =1.$$
Thus, if $S_{n}/a_n \rightarrow 0$, then
$$ \frac{X_n}{a_n} = \frac{S_n}{a_n} - \frac{a_{n-1}}{a_n} \frac{S_{n-1}}{a_{n-1}} \rightarrow 0,$$
since $a_{n-1}/a_n \leq 1$ by monotonicity of the sequence $(a_n)_n$. Therefore $\{S_n/a_n \rightarrow 0 \}$ is a nullset.
