Definition
The sequence $a_n = o(x_n)$ if $a_n/x_n \to 0$. We would read it as $a_n$ is of smaller order than $1/n$, or $a_n$ is little-oh of $1/n$. In your case, if some term $a_n$ is $o(1/n)$ that means that $n a_n \to 0$. A few examples of sequences that are $o(1/n)$ are $c/n^p$ where $p > 1$, $1/(n\log(n))$, and $1/n^2 + 1/n^3$. Even though writing $o(1/n)$ conveys less information than writing the specific sequence, it takes up less space than writing the whole thing out, and tells us that, in this case, the term goes to $0$ faster than $1/n$ (it could also tell us about the speed at which sequences approach infinity).
Some Other Things
On the other hand, writing $a_n = O(1/n)$ means that $|na_n| < M < \infty$ for $n$ bigger than some $n_0$. Typically (but not always) this means that $a_n = c/n$. This means that $a_n$ is of the same or smaller order than $1/n$.
Another thing that might pop up is when people write $o_p(x_n)$ or $O_p(x_n)$. Replace all the definitions above with convergence in probability. It doesn't look like you're using this, however, because you aren't talking about convergence of random variables...just their mgfs.
Your Problem
Also, are you sure that those are what your mgfs look like multiplied together? Specifically the part $1 + \sum_{i=0}^{\infty} \frac{(t/n)^i}{i!}$. I suspect you mean $\sum_{i=0}^{\infty} \frac{(t/n)^i}{i!} = e^{t/n}$ because
\begin{align*}
M_{\bar{X}}(t) &= E[e^{t/n \sum_iX_i}] \\
&= \prod_{i} M_{X_i}(t/n) \\
&= [(1-p)+pe^{t/n}]^n \\
&= \left[(1-p) + p\left\{ \sum_{i=0}^{\infty} \frac{(t/n)^i}{i!}\right\} \right]^n \\
&= \left[ 1 - p + p\left\{ 1 + t/n + \sum_{i=2}^{\infty}\frac{ (t/n)^{i} }{i!} \right\}\right]^n \\
&= \left[1 + pt/n + \sum_{i=2}^{\infty} \frac{pt^i}{i!n^i} \right]^n
\end{align*}
Your professor means $\sum_{i=2}^{\infty} \frac{pt^i}{i!n^i} = o(1/n)$ because
$$
\sum_{i=2}^{\infty} \frac{pt^i}{i!n^{i-1}} \to 0
$$
as $n \to \infty$ as $i-2 \ge 1$.
