Durrett, probability theory and examples exercise 2.7,5 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?
 A: Here is an answer to the second part of your question; I will try to find time tomorrow to address the first part.
The equality
$$
\tag{*}
P(S_n\geq na)\;\geq\; P(S_{n-1}\geq -n\epsilon)\;P(X_n\geq n(a+\epsilon))$$
follows from the inclusion
$$
\left\{S_n\geq na\right\}\supset \{S_{n-1}\geq -n\epsilon\}\cap \{X_n\geq n(a+\epsilon)\}
$$
and independence between $S_{n-1}$ and $X_n$. Then applying $\log$ and multiplying by $-1/n$ shows that
$$
0\leqslant -\frac 1n\log P(S_n\geq na)\leqslant -\frac{1}{n}\log P(S_{n-1}\geq -n\epsilon)
-\frac 1n\log P(X_n\geq n(a+\epsilon)).
$$
Since the random variables $X_i$ are centered  and identically distributec, $S_n/(n-1)\to 0$ in probability hence $P(S_{n-1}\geq -n\epsilon)\to 1$. Moreover, $X_n$ has the same distribution as $X_1$ hence it suffices to check that
$$
\frac 1n\log P(X_1\geq n(a+\epsilon))\to 0.
$$
A: For the first part, if the lim sup was $<0$, we could pick $\beta>0$ such that for $n$ large enough
$$
\frac{1}{n}\log P(X_n>na)\le-\beta 
$$
Then
$$
\begin{split}
\int e^{\theta x}dF
&\le P(X_i<0)+\sum_{n=0}^\infty\int_{na\le X_i\le (n+1)a}e^{\theta x}dF\\
&\le 1+\sum_{n=0}^{N-1}\int_{na\le X_i\le (n+1)a}e^{\theta x}dF+\sum_{n=N}^\infty\int_{na\le X_i\le (n+1)a}e^{\theta x}dF\\
&\le C+\sum_{n=N}^\infty e^{a(n+1)\theta}\cdot e^{-\beta n}\\
&\le C+\sum_{n=N}^\infty e^{-n(\beta-a\theta)+a\theta}
\end{split}\tag{$*$}
$$
Here $C$ and $N$ are constants. \
We can pick $\theta>0$ so small that
$$
\beta-a\theta>0
$$
and thus the last term  of $(*)$ would converge, which is a contradiction to $Ee^{\theta X_i}=\infty$.
