Let $(X_n)_{n \in \mathbb{N}}$ be a sequence of indepent random variables all uniformly distributed on $[-1,1]$
Calculate the probability that
$$\sum_{n=1}^{\infty}{\frac{X_n}{n}}$$ converges
Using Kolmogorov's 0-1 law I know that the probability is either $0$ or $1$. I think that the probability is $1$.
This is my attempt
Using summation by parts we get
$$\sum_{k=1}^{n}{\frac{X_k}{k}} = \frac{S_n}{n} + \sum_{k=1}^{n-1}{S_k \frac{1}{n(n+1)}}$$
Where $$S_i := X_1 + ... + X_{i-1} + X_i$$
Using the strong law of large numbers I get that
$$P\bigg(\lim_{n\to\infty}{\frac{S_n}{n}} = 0\bigg) = 1$$
So basically I can assume that $\lim_{n \to \infty}{\frac{S_n}{n}} = 0$, substituting this in the previous equation gives
$$\sum_{n=1}^{\infty}{\frac{X_n}{n}} = \sum_{n=1}^{\infty}{\frac{S_n}{n(n+1)}}$$
So the probability that $\sum_{n=1}^{\infty}{\frac{X_n}{n}}$ converges is the same as the probability that $\sum_{n=1}^{\infty}{\frac{S_n}{n(n+1)}}$ converges.
Now clearly if $\limsup_{n \to \infty}{\frac{|S_n|}{\sqrt{n}}} < +\infty$ then $\sum_{n=1}^{\infty}{\frac{S_n}{n(n+1)}}$ converges, therefore my idea is to prove that
$$P\bigg(\limsup_{n \to \infty}{\frac{|S_n|}{\sqrt{n}}} < +\infty\bigg) = 1 $$
But I don't know how to do that.
More in general, if $(a_n)_{n \in \mathbb{N}}$ is a sequence of strictly positive real numbers such that $\sum_{n=1}^{\infty}{a_n} < \infty$ then it's sufficient to prove that
$$P\bigg(\limsup_{n \to \infty}{\frac{ |S_n|}{n^2a_n}} < +\infty\bigg) = 1$$
The series converges with probability $1$ by a simple application of Kolmogorov's Three Series Theorem (take $A = 1$ )