Proof that series converges in probability. Consider $X_{1} \dots X_{n} \dots$ - independent random variables with Cauchy-distribution with scale parameter $\theta_{n}$ and zero location parameter such as : $\sum_{n \ge 1} \theta_{n}$ converges. Prove that $X = \sum_{n \ge 1} X_{n}$ converges in probability.
My attempt : let's try to use Cauchy-criteria 
$\forall \epsilon , \delta \exists N $: $\forall n, m \> N $ we have $\operatorname{\mathbb{P}}(| \sum_{n \le k \le m} X_{k}| \ge \epsilon) \le \delta$. Now let's use Markov's inequality $LHS \le \frac{\sum \operatorname{\mathbb{E}(X_{k})}}{\epsilon}$. And I thought that it's converges to zero. But I know there isn't exist a expected value of Cauchy random variable. 
Also I thought about : I can prove that sum of Cauchy variables with parameters $\{\theta_{n}\}$ is a random variable with parameter to be equals $\theta = \sum \theta_{n}$. So I have a random variable with Cauchy distribution. But how can it help me ?
Any advices ?

By Cauchy criterion for convergence in probability it suffices to prove that $$\lim_{n,m\to \infty}P\left(\left|\sum_{k=n+1}^m X_k \right| > \epsilon \right) = 0$$
A simple computation with characteristic functions proves that $\sum_{k=n+1}^m X_k$ follows a Cauchy distribution with location $0$ and scale parameter $\sum_{k=n+1}^m \theta_k$. Since this distribution is symmetric, $$P\left(\left|\sum_{k=n+1}^m X_k \right| > \epsilon \right) = 2P\left(\sum_{k=n+1}^m X_k  > \epsilon \right)=2(1-\frac 1\pi \arctan\left(\frac{\epsilon}{\sum_{k=n+1}^m \theta_k}\right)-\frac 12)$$
By the original assumption, $\lim_{n,m\to \infty}\sum_{k=n+1}^m \theta_k = 0$, thus $\lim_{n,m\to \infty}\frac 1\pi \arctan\left(\frac{\epsilon}{\sum_{k=n+1}^m \theta_k}\right) = \frac 12$ and the result follows.
