<s>[INST] Convergence to zero I need to prove that if $n \rightarrow \infty$ then this sum converges to zero.
$$
 \sum_{k = [\frac{n}{2}]}^n
 {n \choose k}
 Q^k
 \cdot
 \left(
  1-Q
 \right)^{n-k}
$$
In this case $Q$ is constant and equals $e^{-1}$. But it seems that if this is true then it can be generalized for $Q < \frac{1}{2}$.
 [/INST] What you write is the probability of $P(\sum_{k=1}^nX_n \geq \lfloor \frac{n}{2}\rfloor)$, where $X_i$'s are i.i.d Bernoulli variable with parameter $Q$.
$P(\sum_{k=1}^nX_n \geq \lfloor \frac{n}{2}\rfloor) = P(\dfrac{\sum_{k=1}^nX_n -nQ}{\sqrt{n}} \geq \dfrac{\lfloor \frac{n}{2}\rfloor -nQ}{\sqrt{n}})$ and
\begin{align}
\lim_{n\to \infty} \dfrac{\lfloor \frac{n}{2}\rfloor -nQ}{\sqrt{n}} = \begin{cases} +\infty & Q < \frac{1}{2}\\ 0 & Q = \frac{1}{2} \\ -\infty & Q > \frac{1}{2}\end{cases}
\end{align}
So by central limit theorem, your sum converges to $P( G \geq \lim_{n\to \infty} \dfrac{\lfloor \frac{n}{2}\rfloor -nQ}{\sqrt{n}})$, where $G$ is a centred normal distribution with variance $Q(1-Q)$. In particular we have 
\begin{align}
P( G \geq \lim_{n\to \infty} \dfrac{\lfloor \frac{n}{2}\rfloor -nQ}{\sqrt{n}}) = \begin{cases} 0 & Q < \frac{1}{2}\\ \frac{1}{2} & Q = \frac{1}{2} \\ 1 & Q > \frac{1}{2}\end{cases}
\end{align}
</s>