Asymptotic of $\sum_{k=0}^\infty\frac{\Gamma (k+n+1) \Gamma (3 k+n+1)}{\Gamma (k+2) \Gamma (3 k+2 n+2)}$ as $n\to \infty$

Question

Let $a,b,c,d$ be fixed and in a small neighborhood of $0$, so the series $$f(n) = \sum_{k=0}^\infty\frac{\Gamma (a+k+n+1) \Gamma (b+3 k+n+1)}{\Gamma (c+k+2) \Gamma (d+3 k+2 n+2)}$$ converges, the title is special case $a=b=c=d=0$.

Question: What is the leading asymptotic of $f(n)$ as $n\to \infty$, with general $a,b,c,d$ near $0$.

The expression of $f(n)$ is a hypergeometric series and can be converted into an integral form, then a similar analysis like here should be feasible. But this approach is quite complicated, I'm looking for alternative solutions. Thank you very much.

Answer 1

If you simplify the hypergeometric function $$f(n)=\frac{\sqrt \pi}{4^{n-1} }\,\frac{\Gamma (n-2)}{\Gamma \left(n-\frac{1}{2}\right)} \,(A-1)$$ where $$A=\, _4F_3\left(\frac{n-2}{3},\frac{n-1}{3},\frac {n}{3},n;\frac{2 n-1}{3},\frac{2n+1}{3},\frac{2 n}{3};1\right)$$ which is not the most pleasant to work.

A bit of numerical analysis shows that $$\log\Bigg(\frac {f(n)}{4 \sqrt \pi}\Bigg)=a+b \,n +c \log(n)$$

A quick and dirty linear regression gives $(R^2=0.9999954)$

$$\begin{array}{|llll} \hline \text{} & \text{Estimate} & \text{Std Error} & \text{Confidence Interval} \\ a & -1.35909 & 0.05189 & \{-1.46811,-1.25006\} \\ b & -1.13160 & 0.00264 & \{-1.13715,-1.12605\} \\ c & -1.16383 & 0.03442 & \{-1.23615,-1.09151\} \\ \end{array}$$

Using the $ISC$ to produce "nice" numbers $$a \sim -\frac{3+e^{\frac{1}{\pi }}}{5-e^{\gamma }}$$ for an absolute error of $7.33\times 10^{-8}$.

$$b \sim -\frac{3 \left(2+\Gamma \left(\frac{11}{24}\right)\right)}{1+3 \pi }$$ for an absolute error of $1.35\times 10^{-8}$.

$$c \sim -\frac {e-\zeta (5) } {e^{\frac{1}{e}} }$$ for an absolute error of $1.70\times 10^{-7}$.

Answer 2

First of all we note that $$g(n)=\sum_{k=0}^{n}\frac{\Gamma (a+k+n+1) \Gamma (b+3 k+n+1)}{\Gamma (c+k+2) \Gamma (d+3 k+2 n+2)}\ll\sum_{k=n}^\infty\frac{\Gamma (a+k+n+1) \Gamma (b+3 k+n+1)}{\Gamma (c+k+2) \Gamma (d+3 k+2 n+2)}=:S_n$$ at $n\to\infty$

Indeed, using the asymptotic of digamma-function, we see that $$\frac{d}{dk}\frac{\Gamma (a+k+n+1) \Gamma (b+3 k+n+1)}{\Gamma (c+k+2) \Gamma (d+3 k+2 n+2)}$$ $$=\frac{\Gamma (a+k+n+1) \Gamma (b+3 k+n+1)}{\Gamma (c+k+2) \Gamma (d+3 k+2 n+2)}\Big(\psi(a+k+n+1)+\psi(b+3k+n+1)-\psi(c+k+2)-\psi(d+3k+2n+2)\Big)$$ is positive at $k\in[0;n]$; the function reaches its maximum at $k=n$, getting the multiplier $\big(\frac45\big)^{5n}$ in this point. We will se below that the second part of the sum is much bigger. Also, it is straightforward to confirm it by a numeric check.

Supposing $n, k\gg1$ and using the Stirling's asymptotic of gamma-function $$S_n\sim\sum_{k=n}^\infty\sqrt{\frac{(k+n)(3k+n)}{k(3k+2n)}}\frac{(k+n)^a(3k+n)^b}{k^{c+1}(3k+2n)^{d+1}}e^{(k+n)\ln(k+n)+(3k+n)\ln(3k+n)-k\ln k-(3k+2n)\ln(3k+2n)}$$ Using the Euler-Maclauring summation formula we can switch to integration: all other terms of the summation formula provide additional suppression factors at $n, k\gg1$. Making also the substitution $k=nx$ (and in the next line the substitution $t=\frac1x$) $$S_n\sim n^{a+b-c-d-1}\times$$ $$\int_{1}^\infty\frac{(x+1)^{a+1/2}(3x+1)^{b+1/2}}{x^{c+3/2}(3x+2)^{d+3/2}}e^{n\big((x+1)\ln(x+1)+(3x+1)\ln(3x+1)-x\ln x-(3x+2)\ln(3x+2)\big)}dx$$ It is easy to check that the integral converges at $a+b-c-d<1$. After straightforward transformations we get $$S_n\sim n^{a+b-c-d-1}\times$$ $$\int_0^1\frac{(t+1)^{a+1/2}(3+t)^{b+1/2}}{(3+2t)^{d+3/2}}t^{c+d-a-b}e^{n\big((1/t+1)\ln(t+1)+(3/t+1)\ln(3+t)-(3/t+2)\ln(3+2t)\big)}dt$$ The function $h(t)=(1/t+1)\ln(t+1)+(3/t+1)\ln(3+t)-(3/t+2)\ln(3+2t)$ has a single maximum point at $t=0$; near this point $$h(t)=-\ln3-\frac{t^2}{27}+\frac{4t^3}{27}+O(t^4)$$ Using the Laplace' method, $$S_n\sim n^{a+b-c-d-1}\left(\frac13\right)^n\int_0^1\frac{(t+1)^{a+1/2}(3+t)^{b+1/2}}{(3+2t)^{d+3/2}}t^{c+d-a-b}e^{-\frac{nt^2}{27}}dt$$ We see that the main factor is indeed $\left(\frac13\right)^n\gg\Big(\big(\frac45\big)^{5}\Big)^n=\big(0.3276...\big)^n$. Performing integration and keeping only the leading at $n\to\infty$ term: $$\boxed{\,f(n)\sim\left(\frac13\right)^nn^{\frac32(a+b-c-d-1)}3^{\frac32(c-a)+\frac12(d-b)}\frac{\sqrt3}2\Gamma\big(\frac12+\frac{c+d-a-b}2\big)\,}; \,a+b-c-d<1$$ It is straightforward to show that the next asymptotic term has an additional suppression factor $\frac1{\sqrt n}$. Using the third term in $h(t)$ it is not difficult to get the condition, when the approximation is applicable. Supposing that this term should give only small contribution, $$\int_0^1e^{-\frac{nt^2}{27}}\left(1+\frac{4nt^3}{27}\right)dt\approx\sqrt{\frac{27}n}\left(\frac{\sqrt\pi}2+\frac{2\sqrt{27}}{\sqrt n}\right)\,\,\Rightarrow\,\,\sqrt n\gg\frac{4\sqrt{27}}{\sqrt\pi}$$ It means that at least it should be $n\gg100$ - to get a sound approximation. But at $n\gg100 \,\,f(n)$ becomes very small, and WA (free option) refuses to evaluate the sum and integral accurately. That was the reason why I could not obtain a strong numeric check (with the accuracy of several percent), though roughly the result looked correct at any $n$ :)