Stirling's approximation for $ \sum _{k=1}^{n/2} \frac{n!}{k!k!(n-2k)!}a^kb^kc^{n-2k}$

Question

Suppose $a$, $b$, and $c$ are positive real numbers satisfying $a+b+c=1$. I am trying to use Stirling's approximation to obtain an asymptotic (computable) formula for

$$ \sum _{k=1}^{n/2} \frac{n!}{k!k!(n-2k)!}a^kb^kc^{n-2k}$$

as $n \to \infty$. If $n$ is odd then the upper limit of the sum should actually be $(n-1)/2$.

The problem is how to deal with the denominator, seeing as when $n\to \infty$, as far as I understand it, you can't really use asymptotic formulas for $k!$, since $k$, for instance, will take low values for some terms in the summation.

Answer 1

Because $$ \sum _{k=1}^{n/2} \frac{n!}{k!k!(n-2k)!}a^kb^kc^{n-2k}= c^n \sum _{k=1}^{n/2}{n\choose k}{n-k\choose n-2k} \left(\frac{ab}{c^2} \right)^k$$

we want to study the asymptotic behaviour of the sum with term

$$g(k) = {n\choose k}{n-k\choose n-2k} d^k \tag{1}$$

for $0<k<\frac{n}{2}$ and $d=a b /c^2>0$

The first difference $g_1(k)=g(k+1)-g(k)$ is

$$g_1(k)=g(k) \left(\frac{(n-2k)(n-2k-1)}{(k+1)^2} d -1\right) \tag{2}$$

Equating to zero, we get the "critical" index which corresponds to the maximum of $g(k)$

$$k_0=n \frac{1}{2+1/\sqrt{d}} + \frac{\sqrt{d}+2}{4\sqrt{d}+2} + O(n^{-1}) \approx n \frac{1}{2+1/\sqrt{d}} \tag{3} $$

In the same way, we can compute the second difference

$$ \frac{g_2(k)}{g(k)} = \frac{\left( n-2k\right) \,\left( n-2k-1\right) d}{{{\left( k+1\right) }^{2}}}-2+\frac{{{k}^{2}}}{d\,\left( n-2k+1\right) \,\left( n-2k+2\right) } \triangleq h(n,k,d) \tag{4}$$

Now, we want

$$S(n,d)\triangleq \sum_k g(k) \approx \int g(x) dx \tag{5}$$

To get the asymptotics we write the above integral in a form apt for Laplace method:

$$ \int e^{n f(x)} dx \tag{6}$$ with $$f(x)=\frac{\log(g(x))}{n}$$

Laplace method is not really justified her, because $f(x)$ depends on $n$... but we hope it works anyway ($f(x)$ depends rather weakly on $n$). Then we get

$$S(n,d) \approx \sqrt {\frac{2\pi}{n |f''(x_0)|}} e^{n f(x_0)} = \sqrt {\frac{2\pi}{|g''(k_0)/g(k_0)|}} g(k_0) \tag{7}$$

where the last equality comes from $$f''(x)=\frac{1}{n}\left( \frac{g g'' -(g´)^2 }{g^2}\right) \implies f''(x_0)= \frac{1}{n} \frac{g''(x_0)}{g(x_0)} $$

Now, using Stirling approximation:

$$ g(k)\approx \frac{n^n}{k^{2k}(n-2k)^{n-2k} 2 \pi } \sqrt{\frac{n}{k^2(n-2k)}} d^k \tag{8}$$

So finally

$$S(n,d)\approx \sqrt {\frac{2\pi}{|h(n,k_0,d)|} } \frac{n^n}{2 \pi k_0^{2k_0}(n-2k_0)^{n-2k_0} } \sqrt{\frac{n}{k_0^2(n-2k_0)}} d^{k_0} \tag{9}$$

where $k_0$ is given by $(3)$ and $h(n,k_0,d)$ by $(4)$

This gives a good approximation, especially for large $n$.

Alternatively, $h(n,k_0,d)$ can be replaced by the first term of its asympotic expansion (in $n^{-1}$) :

$$h(n,k_0,d)\approx -\frac{2}{n}(4 \sqrt{d}+4+\frac{1}{\sqrt{d}}) \tag{10}$$

For some strange reason (fortuitous cancelling I guess), this approximation works numerically better.

Some numerical examples below - "exact" value corresponds to $\log(S(n,d))$. Approximation $1$ corresponds to $(9)$ (with $(3)$ and $(4)$), approx 2 uses $(10)$.

n        d       exact          approx 1        approx 2
--------------------------------------------------------
10      0.01    0.693147        1.130952        0.648864
50      0.01    7.145196        7.279219        7.137008
400     0.01    69.91077        69.92988        69.90983        
10      0.50    6.994850        7.243550        7.010905        
50      0.50    41.45789        41.51482        41.46112        
400     0.50    348.9017        348.9091        348.9021        
10      1.00    9.099744        9.362582        9.118624        
50      1.00    52.25464        52.31559        52.25839        
400     1.00    435.7325        435.7404        435.7329        
10      10.0    17.92682        18.29466        17.91549        
50      10.0    96.75546        96.86896        96.76006        
400     10.0    792.6511        792.6670        792.6517

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[8px,#ffe]{\ds{\sum _{k = 1}^{n/2}{n! \over k!\,k!\pars{n - 2k}!}\,a^{k}b^{k}c^{n - 2k}}} = -c^{n} + \sum _{k = 0}^{\infty}{n! \over k!\,k!\pars{n - 2k}!}\,a^{k}b^{k}c^{n - 2k} \\[5mm] & = -c^{n} + \sum _{k, p, q\ \in\ \mathbb{N}_{\ \geq\ 0}} {n! \over k!\,p!\,q!}\,a^{k}b^{p}c^{q}\bracks{k + p + q = n}[p = k] \\[5mm] & = -c^{n} + \sum _{k, p, q\ \in\ \mathbb{N}_{\ \geq\ 0}} {n \choose k,p,q}\,a^{k}b^{p}c^{q} \bracks{k + p + q = n}\braces{\vphantom{\Large A}\bracks{z^{0}}z^{p - k}} \\[5mm] & = -c^{n} + \bracks{z^{0}}\sum _{k, p, q\ \in\ \mathbb{N}_{\ \geq\ 0}} {n \choose k,p,q}\pars{a \over z}^{k}\pars{bz}^{\,p}c^{q} \bracks{k + p + q = n} \\[5mm] & = -c^{n} + \bracks{z^{0}}\pars{{a \over z} + bz + c}^{n} = -c^{n} + \bracks{z^{0}} {\pars{bz^{2} + cz + a}^{n} \over z^{n}} \\[5mm] & = -c^{n} + b^{n} \bracks{z^{n}}\pars{z^{2} + {c \over b}\,z + {a \over b}}^{n} = \bbx{\ds{-c^{n} + b^{n} \bracks{z^{n}}\braces{\vphantom{\Large A}\pars{z - r_{-}}^{n} \pars{z - r_{+}}^{n}}}} \end{align}

where $\ds{\braces{r_{-},r_{+}}}$ are the roots of $\ds{z^{2} + {c \over b}\,z + {a \over b} = 0}$. Namely, $$ r_{\pm} = {-c \pm \,\mrm{sgn}\pars{b}\root{c^{2} - 4a^{2}} \over 2b} $$

Further progress is available whenever particular values of $\ds{a,b,c}$ are known.