Computing large deviations probabilities for sums of gaussians 
Let $R_i$ be i.i.d. normally distributed with $\mu = 1$ and $\sigma^2 = 0,1$ and $S_n = \sum_{i=1}^n R_i$ and let $c \geq E[R_i]$.
  What is the probability that $P(S_n > nc)$ for large $n$?

I've thought of using the result that for $X_i$ i.i.d.with expectation $0$ and $Z_n = \sum X_i$ and $M(t)$ the m.g.f. of $X_i$ it follows that
$$P(Z_n > na)^{\frac{1}{n}} \rightarrow \exp(-\psi(a))$$
where $\psi(a) = -\ln(\inf \{\exp(-at)M(t)\})$.
But how can I apply this? Do I have to standardize $R_i$ into $\frac{R_i -1}{0,1}$, then the m.g.f. of this would be $M(t) = \exp(\frac{1}{2} t^2)$. How should I proceed here?
 A: Since $\mathbb{E}(R_i) = \mu = 1$, the centered random variables $X_i := R_i -1$ have mean $0$. Moreover, we have for all $n \in \mathbb{N}$
$$\mathbb{P}(S_n > nc) = \mathbb{P} \left( \sum_{i=1}^n (R_i-1) > n (c-1) \right).$$
Applying the large deviation result mentioned in your question, we obtain
$$\mathbb{P}(S_n>nc)^{1/n} \xrightarrow[]{n \to \infty} \exp(-\psi(c-1)). \tag{1}$$
It remains to identify $\psi$. To this end, note that the moment generating function of the Gaussian random variable $X_i = R_i-1$ is given by $M(t) = e^{t^2 \sigma^2/2}$. Hence,
$$\psi(c) = -\log \left( \inf_t (e^{-ct} e^{t^2 \sigma^2/2}) \right).$$
Differentiating $t \mapsto e^{-ct + t^2 \sigma^2/2}$ it is not difficult to see that the infimum is attained at $t_0 = c/\sigma^2$; hence,
$$\psi(c) =- \log \left( \exp \left[ -\frac{c^2}{2\sigma^2} \right] \right) =  \frac{c^2}{2\sigma^2}.$$
Thus, by $(1)$,
$$\mathbb{P}(S_n>nc)^{1/n} \xrightarrow[]{n \to \infty} \exp \left( \frac{-(c-1)^2}{2\sigma^2} \right)$$
for $c \geq\mu = 1$, i.e.
$$\mathbb{P}(S_n>nc) \approx \exp \left(- n \frac{(c-1)^2}{2\sigma^2} \right)$$
for large $n$.
