Show that $\sum_{k=1}^{n/2} \binom{n}{k}\alpha^{k(n-k)} \rightarrow 0$ for $\alpha <1$ The problem: 
Show that, as $n\rightarrow + \infty$,  $\sum_{k=1}^{n/2} \dbinom{n}{k}\alpha^{k(n-k)} \rightarrow 0$ for $0<\alpha<1$.
What I tried :
I tried to use $\dbinom{n}{k} \leq \dbinom{n}{\lfloor{n/2}\rfloor}$
After that : 
$$
\begin{align}
\sum_{k=1}^{n/2} \dbinom{n}{k}\alpha^{k(n-k)}& \leq \dbinom{n}{\lfloor{n/2}\rfloor}\sum_{k=1}^{n/2}\alpha^{k(n-k)} \\
& \leq \dbinom{n}{\lfloor{n/2}\rfloor}\sum_{k=1}^{n/2}\alpha^k && \text{(cause } \alpha<1) \\
& \leq \dbinom{n}{\lfloor{n/2}\rfloor}\dfrac{1-\alpha^{\lfloor n/2\rfloor}}{1-\alpha}
\end{align}$$
but it is not good enough because $\dbinom{n}{\lfloor{n/2}\rfloor}\dfrac{1-\alpha^{\lfloor n/2\rfloor}}{1-\alpha} \rightarrow +\infty$
 A: Fix $1\leq k_0\leq n/2$ so that $\alpha^{k_0} < 1/2$ and split the sum as
$$
\sum_{k=1}^{n/2}  \dbinom{n}{k} \alpha^{k(n-k)} = \sum_{k=1}^{k_0} + \sum_{k = k_0+1}^{n/2} := A + B,
$$
we will estimate $A$ and $B$ separately.
First observe that 
$$
A \leq \alpha^{n-1} \sum_{k=1}^{k_0} \dbinom{n}{k} \leq \alpha^{n-1} n^{k_0},
$$
where we used a crude estimate $\dbinom{n}{k} = \frac{n(n-1)...(n-k_0 +1)}{k_0!} \leq n^{k_0}$. Since $k_0$ is fixed and $\alpha<1$ we get
$$
A \to 0 \text{ as } n \to \infty.
$$
We now proceed to the second part $B$. Consider the function $f(x) = x (n-x)$, where $x\geq 1$. Since $f'(x) = n - 2x $, we see that $f$ increases in the range $1\leq x \leq n/2$. Hence we have
$k(n-k) \geq k_0(n-k_0)$ for all $k=k_0+1,...,n/2$, from which we obtain
$$
B \leq \sum_{k= k_0 + 1}^{n/2} \dbinom{n}{k}  \alpha^{k_0(n-k_0)}  \leq \alpha^{k_0(n-k_0)} \sum_{k=0}^n \dbinom{n}{k}  = \alpha^{k_0(n-k_0)} 2^n = (2\alpha^{k_0})^n \alpha^{-k_0^2}.
$$
But recall, that $k_0$ was fixed so that $2\alpha^{k_0}<1$, and hence $B \to 0$ as $n\to \infty$.
We thus proved that both $A$ and $B$ converge to $0$, hence so does the original sum.
