Let $b>a>0$. Given $ \lambda>0$, what is the asymptotic behavior of $$F(\lambda):=\sum_{k=0}^{\infty}\frac{\Gamma{(a k)}}{\Gamma{(b k)}}{\lambda}^{-k}$$ as $\lambda\to 0^{+}$ and as $\lambda \to+\infty $ ?

One can try to understand this sum using the well-known Stirling's approximation https://en.wikipedia.org/wiki/Stirling%27s_approximation
$$\Gamma(x)\sim\sqrt{\frac{2\pi}{x}}\left(\frac{x}{e}\right)^{x},\quad\text{as}\;\; x \to +\infty.$$

Using Stirling's formula gives the sum

$$G(\lambda):=\sqrt{\frac{b}{a}} \sum_{k>>1}^{\infty} \left(\frac{a^a}{b^b}e^{b-a}\right)^{k}\frac{1}{k^{(b-a)k}}\frac{1}{\lambda^{k}}.$$

From the latter we see that $F(\lambda)$ is well-defined for every $\lambda>0$.

Now, how obtain the asymptotic behavior of $G(\lambda)$ as $\lambda\to 0^{+}$ or as $\lambda \to +\infty $ ?

  • $\begingroup$ The asymptotics for $\lambda\to\infty$ are simply $1+\frac{\Gamma(a)}{\Gamma(b)}\frac{1}{\lambda} + O(1/\lambda^2)$. $\endgroup$
    – Diffusion
    Commented Apr 30, 2023 at 4:38
  • $\begingroup$ I have no idea how you got that. A hint for a solution would be helpful. $\endgroup$
    – Medo
    Commented Apr 30, 2023 at 7:06
  • $\begingroup$ @Zachary's answer is true for any convergent power series. But of course $\lambda\to 0$ is not so easy. $\endgroup$ Commented Apr 30, 2023 at 13:39
  • $\begingroup$ @Iosif Pinelis Yes. corrected. Thank you. And thank you for the continuous and generous contributions to mathoverflow. I am reading your answer carefully now. $\endgroup$
    – Medo
    Commented Apr 30, 2023 at 21:55

1 Answer 1


$\newcommand{\Ga}{\Gamma}$The value of \begin{equation*} r_k:=\frac{\Ga(ak)}{\Ga(bk)} \end{equation*} is undefined at $k=0$. So, let \begin{equation*} r_0:=\lim_{k\to0}\frac{\Ga(ak)}{\Ga(bk)}=\frac ba. \end{equation*}

Letting also $x:=1/\lambda$, we see that we need to find the asymptotics of \begin{equation*} f(x):=\sum_{k=0}^\infty t_k\quad\text{with}\quad t_k:=t_k(x):=r_k x^k \tag{10}\label{10} \end{equation*} as $x\downarrow0$ and as $x\to\infty$.

The case $x\downarrow0$ is easy. Indeed, the $r_k$'s are bounded and hence \begin{equation*} f(x)\to r_0=\frac ba \end{equation*} as $x\downarrow0$.

Consider now the case when $x\to\infty$. Then $f(x)\ge t_m=r_m x^m$ for each natural $m$. So, for any function $x\mapsto K(x)$ such that $K(x)$ increases to $\infty$ slowly enough as $x$ increases to $\infty$, we have
\begin{equation*} f(x)\sim\sum_{k\ge K(x)} t_k. \tag{20}\label{20} \end{equation*} By Stirling's formula,
\begin{equation*} t_k\sim c_1 e^{g(k)} \tag{30}\label{30} \end{equation*} as $k\to\infty$, where \begin{equation*} g(v):=g_y(v):=v\ln y-cv\ln v, \end{equation*} \begin{equation*} c_1:=\sqrt{\frac ba},\quad y:=c_2x\to\infty,\quad c_2:=\frac{(a/e)^a}{(b/e)^b},\quad c:=b-a. \tag{35}\label{35} \end{equation*} The concave function $g$ increases on $(0,v_y]$ and decreases on $[v_y,\infty)$, where $$v_y:=y^{1/c}/e.$$ So, \begin{equation*} \Big|\sum_{k\ge0} e^{g(k)}-I\Big|\le2e^{g(v_y)}, \tag{40}\label{40} \end{equation*} where \begin{equation*} I:=I(y):=\int_0^\infty e^{g(v)}\,dv. \end{equation*} Note that
\begin{equation*} g(v_y)=cv_y=cy^{1/c}/e\to\infty. \end{equation*}

Write \begin{equation*} I=I_1+I_2+I_3, \tag{50}\label{50} \end{equation*} where \begin{equation*} I_1:=\int_0^{v_y-v_y^{2/3}} e^{g(v)}\,dv,\quad I_2:=\int_{v_y-v_y^{2/3}}^{v_y+v_y^{2/3}} e^{g(v)}\,dv,\quad I_3:=\int_{v_y+v_y^{2/3}}^\infty e^{g(v)}\,dv. \end{equation*} We have $g'(v)=\ln y-c-c\ln v$, $g'(v_y)=0$, and $g''(v)=-c/v\sim-c/v_y$ for $v\sim v_y$ and hence for $v\in[v_y-v_y^{2/3},v_y+v_y^{2/3}]$. So, \begin{equation*} g(v)-g(v_y)=-\frac{c+o(1)}{v_y}\frac{(v-v_y)^2}2\quad\text{for}\quad v\in[v_y-v_y^{2/3},v_y+v_y^{2/3}]. \end{equation*} Using the substitution $z=(v-v_y)/\sqrt{v_y/c}$, we get \begin{equation*} I_2=e^{g(v_y)}\sqrt{\frac{v_y}c}\,\int_{-v_y^{1/6}\sqrt c}^{v_y^{1/6}\sqrt c} e^{-z^2/(2+o(1))}\,dz \sim\sqrt{\frac{2\pi v_y}c}\,e^{g(v_y)}, \tag{55}\label{55} \end{equation*} since $v_y\to\infty$.

It also follows from $g'(v_y)=0$ and $g''(v)=-c/v\sim-c/v_y$ for $v\sim v_y$ that $g'(v_y+v_y^{2/3})\le-v_y^{2/3}(c+o(1))/v_y=-(c+o(1))v_y^{-1/3}$. Therefore and because the function $g$ is concave, for all real $v\ge v_y+v_y^{2/3}$ we have \begin{equation*} \begin{aligned} g(v)&\le g(v_y+v_y^{2/3})+g'(v_y+v_y^{2/3})(v-(v_y+v_y^{2/3})) \\ &\le g(v_y)-\frac{c+o(1)}{v_y}\frac{v_y^{4/3}}2-(c+o(1))v_y^{-1/3}(v-(v_y+v_y^{2/3})) \\ \end{aligned} \end{equation*} So, \begin{equation*} \begin{aligned} I_3&\le e^{g(v_y)} \exp\Big(-\frac{c+o(1)}2v_y^{1/3}\Big) \\ &\times \int_{v_y+v_y^{2/3}}^\infty\exp(-(c+o(1))v_y^{-1/3}(v-(v_y+v_y^{2/3})))\,dv \\ &\sim e^{g(v_y)} \exp\Big(-\frac{c+o(1)}2v_y^{1/3}\Big)\frac{v_y^{1/3}}c =o(e^{g(v_y)})=o(I_2), \end{aligned} \end{equation*} in view of \eqref{55}. Similarly, $I_1=o(I_2)$.

So, by \eqref{50}, \begin{equation*} I\sim I_2\sim \sqrt{\frac{2\pi v_y}c}\,e^{g(v_y)}. \tag{60}\label{60} \end{equation*} So, in view of \eqref{40} and because $v_y\to\infty$, we have \begin{equation*} \sum_{k\ge0} e^{g(k)}\sim I. \tag{70}\label{70} \end{equation*} Similarly to \eqref{20}, \begin{equation*} \sum_{k\ge K(x)} e^{g(k)}\sim\sum_{k\ge0} e^{g(k)} \tag{80}\label{80} \end{equation*} for any function $x\mapsto K(x)$ such that $K(x)$ increases to $\infty$ slowly enough as $x$ increases to $\infty$.

Collecting \eqref{20}, \eqref{30}, \eqref{80}, \eqref{70}, and \eqref{60}, for $x\to\infty$ we get \begin{equation*} f(x)\sim c_1 \sqrt{\frac{2\pi v_y}c}\,e^{g(v_y)} =c_1 \sqrt{\frac{2\pi(c_2 x)^{1/c}}{ec}}\,e^{c(c_2 x)^{1/c}/e}=:\tilde f(x), \tag{90}\label{90} \end{equation*} with the constants $c_1,c_2,c$ as defined in \eqref{35}.

To illustrate (and check) the asymptotic relation \eqref{90}, here is (part of) the graph $\{(x,f(x)/\tilde f(x))\colon0<x\le10\}$ for $a=1$ and $b=2$ (as well as the dotted line $\{(x,1)\colon0<x\le10\}$):

enter image description here


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