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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 . enter image description here

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?

Przemo
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    $\mathfrak{What} ,,\mathfrak{are} ,,\mathfrak{A}_p ,, \mathfrak{and} ,, \mathfrak{B}_p$? – mvw Jul 25 '14 at 17:11
  • These are some numbers that depend on $p$ only. The logarithms of those numbers are equal to the intercept and the slope of the lines in the left Figure. The whole problem is to find those numbers in a closed form. this is the question that I would like to answer. Is there any generic approach to analyzing such problems? – Przemo Jul 25 '14 at 17:16
  • $\mathfrak{Thanks},, \mathfrak{a},, \mathfrak{lot}.\mathfrak{And} ,,\mathfrak{what} ,,\mathfrak{are},,\mathfrak{the},, C_j^n?$ – mvw Jul 25 '14 at 17:25
  • $C^n_j$ = n!/(j!(n-j)!)$ is the binomial factor. – Przemo Jul 25 '14 at 17:26
  • The binomial coefficients are more commonly denoted $\binom{n}{j}$, obtained via \binom{n}{j}. – Daniel Fischer Jul 25 '14 at 17:40

1 Answers1

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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}

Przemo
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