Will Central Limit Theorem work here? suppose that $X_1, X_2, \cdots $ are i.i.d. random variables with mean $0$ and variance $1$. Let $Z \sim N(0,1)$. For some constant $a \in \mathbb{R}$, define sequences
$$ \delta_n = Pr\left(\frac{X_1+X_2+\cdots+X_n}{\sqrt{n}} \leq a \right)$$
$$ \gamma_n = Pr\left(\frac{X_1+X_2+\cdots+X_n}{\sqrt{n}} \leq a + \frac{1}{n} \right)$$ 
Then, the central limit theorem tells us that $\delta_n \to Pr(Z \leq a)$. The proof I am somewhat familiar with uses MGF approach. 
My question is this: What can we say about $\lim_{n \to \infty} \gamma_n$?
More generally, suppose we had $a+ g_n$ where $g_n$ is a sequence that goes to $0$. Is it true that $\gamma_n \to Pr(Z \leq a)$ as well? (Under the same assumptions under which $\delta_n \to Pr(Z \leq a)$ is true)
 A: Even the general case holds as you would hope; that is, if $g_n \to 0$, then $\gamma_n \to Pr(Z \leq a)$.
For convenience, let $S_n := \frac{X_1 + \dots + X_n}{\sqrt n}$. The proof that you know establishes that $S_n \stackrel{d}{\to} Z$; that is,
$$\mathbb P (S_n \leq c) \to \mathbb P (Z \leq c) $$
for any fixed $c$. 
Now, let $g_n \to 0$ and let $\epsilon > 0$ be fixed. We wish to show that for sufficiently large $n$, we have
$$|\mathbb P(S_n \leq c + g_n) - \mathbb P(Z \leq c)| < \epsilon.$$
Since $Z$ is a continuous random variable, the function $\mathbb P(Z \leq c)$ is continuous in $c$; that is, there is some small $\delta > 0$ such that
$$|\mathbb P(Z \leq c + \delta) - \mathbb P(Z \leq c)| < \epsilon/3.$$
Choose $n$ sufficiently large so that $g_n < \delta$. It follows that for sufficiently large $n$,
$$\mathbb P(S_n \leq c) \leq \mathbb P(S_n \leq c + g_n) \leq \mathbb P(S_n \leq c + \delta).$$
As $n \to \infty$, the left side can be assumed to be at least $\mathbb P(Z \leq c) - \epsilon/3$ and the right side can be assumed to be at most $\mathbb P(Z \leq c + \delta) + \epsilon/3 \leq \mathbb P(Z \leq c) + 2 \epsilon/3$. We have therefore bounded $\mathbb P(S_n \leq c + g_n)$ and $\mathbb P(Z \leq c)$ together in an interval of length at most $\epsilon$.
A: Note that $\delta_n \leq \gamma_n$. Also define $$\zeta_n(\epsilon) 
= Pr\left(\frac{X_1+X_2+\cdots+X_n}{\sqrt{n}} \leq a +\epsilon \right)$$ with $\epsilon > 0$. Then, for any such $\epsilon$ there exists an $n_0$ such that when $n\geq n_0$ it follows that $\gamma_n \leq \zeta_n(\epsilon)$.
