Asymptotic behaviour of a sum Let $p$ and $q$ be positive real numbers such that $p+q = 1$. am interested in in the large-$n$ behaviour of a following sum:
\begin{equation}
\sum\limits_{j=0}^{n-1} \left(1 + \frac{n-j-1/2}{j+1} \frac{q}{p}\right) C^{n-1}_j C^n_j q^j
\end{equation}
I suspect that the sum behaves like:
\begin{equation}
{\mathfrak A}_p ({\mathfrak B}_p)^n
\end{equation}
as $n \rightarrow \infty$.
Below I enclose two figures. The one on the left shows the sum as a function of n for different values of $q=0,3,0.4,\cdots,0.9$ (in Red, Orange,Magenta,..,Green and Blue respectively). The one on the right shows a logarithmic derivative of that sum .

As we can see the sum clearly behaves exponentialy for big values of $n$. How do I find the closed form solution that describes that behaviour?
 A: The sum in question consists of two terms. Let us analyze the first term only. We will see later that the second term can be handled in a similar manner. Therefore we have:
\begin{eqnarray}
S &:=& \sum\limits_{j=0}^{n-1} \binom{n-1}{j} \binom{n}{j} q^j \\
  &=&  \sum\limits_{j_1=0}^{n-1}  \sum\limits_{j_2=0}^{n} \binom{n}{j_1} \binom{n}{j_2} \sqrt{q}^{j_1+j_2} \delta_{j_1,j_2} \\
  &=& \int\limits_{-\pi}^{\pi} \left(1 + \sqrt{q} e^{\imath \phi}\right)^{n-1} \left(1 + \sqrt{q} e^{-\imath \phi}\right)^n \frac{d \phi}{2 \pi}
\end{eqnarray}
Here we used the representation of the Kronecker delta function:
\begin{equation}
\delta_{i,j} := \int\limits_{-\pi}^\pi e^{\imath \phi (i-j)} \frac{d \phi}{2 \pi}
\end{equation}
We simplify the result further and get:
\begin{equation}
S = \int\limits_{-\pi}^\pi \exp\left[(n-1/2) \log(1+q + 2\sqrt{q} \cos(\phi))\right] e^{-\imath k(\phi)} \frac{d \phi}{2 \pi}
\end{equation}
where 
\begin{equation}
k(\phi) := \arcsin\left(\frac{\sqrt{q} \sin(\phi)}{\sqrt{1+q + 2 \sqrt{q} \cos(\phi)}}\right)
\end{equation}
Now we expand the term inside the first exponent in a Taylor series about $\phi=0$ and retain terms up to order two only. We have:
\begin{eqnarray}
S &\simeq& \left(1 + \sqrt{q}\right)^{2n-1} \int\limits_{-\pi}^{\pi} e^{-(2n-1) \frac{\sqrt{q}}{(1+\sqrt{q})^2} \frac{\phi^2}{2}} e^{-\imath k(\phi)} \frac{d \phi}{2 \pi} \\
& = & \left(1 + \sqrt{q}\right)^{2n-1} \sqrt{2 \pi \sigma_n^2} \int\limits_{-\pi}^\pi \delta_n(\phi) e^{-\imath k(\phi)} \frac{d\phi}{2 \pi}
\end{eqnarray}
Here we defined:
\begin{equation}
\sigma_n^2 := \frac{(1+\sqrt{q})^2}{(2n-1) \sqrt{q}}
\end{equation}
and we used the representation of a Dirac delta function:
\begin{equation}
\delta_n(\phi) := \frac{1}{\sqrt{2 \pi \sigma_n^2}} e^{-\frac{\phi^2}{2 \sigma_n^2}}
\end{equation}
For big values of $n$ the representation of the Dirac delta function picks out the integrand value at $\phi=0$. Simplifying the whole thing we obtain a neat result:
\begin{equation}
S = \frac{1}{\sqrt{2 \pi \sqrt{q}} }
    \frac{\left(1+\sqrt{q}\right)^{2 n}}{\sqrt{2 n-1}}
\end{equation}
As we can see my conjecture was almost correct. In fact we have:
\begin{equation}
\lim\limits_{n\rightarrow \infty} S_n = {\mathfrak A}_q ({\mathfrak B}_q)^n \cdot \frac{1}{\sqrt{2n-1}}
\end{equation}
where 
\begin{equation}
\left\{ {\mathfrak A}_q,{\mathfrak B}_q \right\} := 
\left\{ \frac{1}{\sqrt{2 \pi \sqrt{q}}}, \left(1+\sqrt{q}\right)^2 \right\}
\end{equation}
