# Question in Lemma 2.4.2 of Rick Durrett's Probability: Theory and Examples version 5

I was studying Durrett's proof of Strong Law of Large numbers and I got stuck while trying to understand the following Lemma:

Let $$X_1,X_2,\dots$$ be pairwise independent identically distributed random variables with $$E|X_i|<\infty$$. Let $$EX_i=\mu$$ and $$S_n=X_1+\dots+X_n$$. Let $$Y_k = X_k1_{(|X_k|\leq k)}$$ and $$T_n = Y_1 + \dots + Y_n$$. To show that $$S_n/n\to\mu$$ a.s., it is sufficient to prove that $$T_n/n \to \mu$$ a.s. as $$n\to\infty$$.

Durrett first shows that $$P(X_k\neq Y_k \text{ i.o.})=0$$. This means that for almost every $$\omega$$, $$|S_n(\omega)-T_n(\omega)|\leq R(\omega)<\infty$$ for all $$n$$. Durrett concludes by saying that from here the desired result follows.

My attempt to complete the proof is as follows:

To show that $$S_n/n \to \mu$$ a.s. it is enough to show that for any $$\epsilon>0$$, $$P(|S_n/n - \mu|>\epsilon \text{ i.o.})=0$$. By the Borel Cantelli Lemma, the above statement is true when $$\sum_{n=1}^{\infty}P(|S_n/n - \mu|>\epsilon)<\infty$$. By using a union bound, $$\sum_{n=1}^{\infty}P(|S_n/n - \mu|>\epsilon)\leq \sum_{n=1}^{\infty} P(|T_n/n - \mu|>\epsilon/2)+\sum_{n=1}^{\infty} P(|S_n-T_n|/n>\epsilon/2)$$. The expression $$P(|T_n/n - \mu|>\epsilon/2)$$ is finite by our assumption.

I can't figure out how to show that the other expression is finite. I know that $$P(|S_n-T_n|/n>\epsilon/2) = P(|R|/n>\epsilon/2)$$ but I don't know what to do from here. How do I show $$\sum_{n=1}^{\infty}P(|R|/n>\epsilon/2)<\infty$$?

Since $$X_n = Y_n$$ for $$n \geq N$$, $$\frac{S_n - T_n}{n} = \frac{S_N - T_N}{n} \to 0 \text{ as }n \to \infty.$$