mean sample convergent to infinity almost surely I'm trying to prove the following exercise:

Use truncated random variables to prove that if $(X_{i})_{i\in\mathbb{N}}$ are random variables independent and identical distributed, with $E(X_{i}^+)=\infty, \,E(X^-)<\infty$ and $S_{n}=X_{1}+\cdots+X_{n}$, then $$\frac{S_{n}}{n}\rightarrow\infty$$ almost surely when $n\rightarrow\infty.$

My attemp is, if we use truncated random variables, then expectation of every $X_{n}\mathbf1_{|X_{n}|\leq n},$ exist because of the boundedness of each truncated random variable. But I don't know how to use that to prove the proposition.
Any kind of help is thanked in advance.

For $k,i=1,2,\dots$ define: $$X_{k,i}:=X_i\mathbf1_{X_{i}\leq k}$$
Moreover for $n=1,2,\dots$ define: $$S_{k,n}=X_{k,1}+\cdots+X_{k,n}$$
Then $\mu_k:=\mathbb EX_{k,1}<\infty$ and:$$\frac{S_{k,n}}{n}\to\mu_k\text{ a.s.}$$
This with $S_n\geq S_{k,n}$ so that: $$\liminf\frac{S_n}{n}\geq\mu_k\text{ a.s.}$$
This for every $k$ so that $\lim_{k\to\infty}\mu_k=+\infty$ allows us to conclude:$$\frac{S_n}{n}\to+\infty\text{ a.s.}$$
