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?
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}