The problem is as below.
Suppose $EX_i=0$ and $E\exp(\theta X_i)=\infty$ for all $\theta>0$. Let $S_n=X_1+\dots+X_n$.
Then $$\frac{1}{n}\log P(S_n \geq na) \rightarrow 0\quad\text{for all}\quad a>0.$$
The solution uses the fact that $$P(S_n\geq na)\;\geq\; P(S_{n-1}\geq -n\epsilon)\;P(X_n\geq n(a+\epsilon)).$$
This is different from the original large deviation proof, where we use $$P(S_n \geq na)\;\geq\; P(S_{n-1} \geq (a-a'+\epsilon)n)\;P(X_{n}\in[(a'-\epsilon)n,(a'+\epsilon)n]).$$
The solution says that $Ee^{\theta X}=\infty$ implies $\limsup_{n\rightarrow\infty}\frac{1}{n}\log P(X_n >na)=0.$
How do you prove this and how does the result follows from this fact?