Pove that the summation of iid sequence satisfies $\frac{S_n}{n\log n}\rightarrow c\quad \text{in probability}$

Suppose that $$X_1,X_2,\cdots,X_n$$ are iid sequence with pdf $$\frac{2}{\pi (1+x^2)}\cdot 1_{(0,+\infty)}(x)$$. Denote $$S_n$$ as $$S_n:=X_1+X_2+\cdots+X_n$$. Prove that there exits $$c>0$$ such that $$\frac{S_n}{n\log n}\rightarrow c\quad \text{in probability}$$ and calculate $$c$$.

My solution: Choose another iid sequence $$Y_1,Y_2,\cdots,Y_n$$ such that $$X_n,Y_n$$ are independent and have same distribution. Let $$X_n^s:=X_n-Y_n$$. I have proved that $$\frac{X_1^s+X_2^s+\cdots+X_n^s}{n\log n}\rightarrow 0\quad \text{in probability}.$$ Therefore, $$\frac{S_n}{n\log n}-m_n\rightarrow 0\quad \text{in probability}$$ where $$m_n$$ is the medin of $$\frac{S_n}{n\log n}$$. I want to show $$m_n\rightarrow c$$ but I can't.

Let $$S_n' = \sum_{k=1}^n X_k 1_{\{|X_k|\le a_n\}}$$. Clearly,
$$\mathbb P\left(\left|\frac{S_n - b_n}{a_n}\right|>\varepsilon\right)\le\mathbb P(S_n\ne S_n') + \mathbb P\left(\left|\frac{S_n' - b_n}{a_n}\right|>\varepsilon\right).$$
With $$a_n = n\log n$$ and $$b_n = \mathbb E S_n'$$, we note that $$\mathbb P(S_n\ne S_n')\le\sum_{k=1}^n\mathbb P(|X_k|>a_n)\to 0,$$ and that $$\mathbb P\left(\left|\frac{S_n' - b_n}{a_n}\right|>\varepsilon\right)\le \frac{1}{\varepsilon^2 a_n^2}\text{Var}\left(S_n'\right)\le\frac{1}{\varepsilon^2 a_n^2}\sum_{k=1}^n\mathbb E(X_k^2 1_{\{|X_k|\le a_n\}})\to 0.$$ So we have $$\frac{S_n - b_n}{a_n}\to 0\quad\text{in probability}$$ with $$b_n/a_n\to 2/\pi$$. Then it's easy to see $$\frac{S_n}{n\log n}\to\frac{2}{\pi}\quad\text{in probability.}$$
Note that $$\frac{S_n}{n\log n}$$ can be divided into two parts: $$\frac{S_n}{n\log n}=\frac{\sum_{k=1}^nX_k 1_{X_k\le n\log n}}{n\log n}+\frac{\sum_{k=1}^nX_k 1_{X_k> n\log n}}{n\log n}$$ The second part converges to zero in probability, and one can verify the mean square of the first part minus $$c$$ converges to $$(2/\pi-c)^2$$. Therefore, $$\frac{S_n}{n\log n}$$ converges $$c=2/\pi$$ in probability.