Evaluating $\lim_{z\to n}\frac{\Gamma (1-z)}{\Gamma (1-2 z)}$ using Mathematica

Question

I am trying to evaluate: $$\lim_{z\to n}\frac{\Gamma (1-z)}{\Gamma (1-2 z)}$$ Here are my steps: \begin{align*} \lim_{z\to n}\frac{\Gamma (1-z)}{\Gamma (1-2 z)}&=\lim_{z\to n}\frac{\Gamma (-n-z+1) \prod _{k=0}^{n-1} (-k-z)}{\Gamma (1-2 z)}\\[4pt] &=\left(\lim_{z\to n}\frac{\Gamma (-n-z+1)}{\Gamma (1-2 z)}\right) \lim_{z\to n}\prod_{k=0}^{n-1}(-k-z) \\[4pt] &=(-1)^n \left(\lim_{z\to n}\frac{\Gamma(-n-z+1)}{\Gamma (1-2 z)}\right) \lim_{z\to n}\prod_{k=0}^{n-1} (k+z) \end{align*} Here is the problem, also verified by Mathematica, $$ \lim_{z\to n}\frac{\Gamma (-n-z+1)}{\Gamma (1-2 z)}=1 $$ so we should expect that \begin{align*} (-1)^n \left(\lim_{z\to n}\frac{\Gamma (-n-z+1)}{\Gamma (1-2 z)}\right) \lim_{z\to n}\prod_{k=0}^{n-1} (k+z)&=(-1)^n \lim_{z\to n}\prod_{k=0}^{n-1} (k+z)\\ &=(-1)^nn(n+1)\cdots(2n-1)\\ &=(-1)^n\frac{1\cdot2\cdots n(n+1)\cdots(2n-1)({\color{red} 2}n)}{{\color{red} 2}\cdot 1\cdot2\cdots n}\\ &=\frac{(-1)^n}{2}\frac{(2n)!}{n!} \end{align*} One may, then, conclude that \begin{align*} \lim_{z\to n}\frac{\Gamma (1-z)}{\Gamma (1-2 z)}=\frac{(-1)^n}{2}\frac{(2n)!}{n!} \end{align*} But Mathematica gives \begin{align*} \lim_{z\to n}\frac{\Gamma (1-z)}{\Gamma (1-2 z)}={(-1)^n}\frac{(2n)!}{n!} \end{align*} Am I missing something here? Of course, there are other approaches, but I want to understand why! there is such inconsistency.

Answer 1

Actually $$\lim_{z\to n}\frac{\Gamma(-n-z+1)}{\Gamma(1-2z)}=2.$$ This can be shown using $$\lim_{z\to-n}(z+n)\Gamma(z)=\color{gray}{\lim_{z\to-n}\frac{\Gamma(z+n+1)}{z(z+1)\cdots(z+n-1)}}=\frac{(-1)^n}{n!}$$ (where $n$ is a nonnegative integer), as well as the original limit: $$\lim_{z\to n}\frac{\Gamma(1-z)}{\Gamma(1-2z)}=2\lim_{z\to n}\frac{\big((1-z)+(n-1)\big)\Gamma(1-z)}{\big((1-2z)+(2n-1)\big)\Gamma(1-2z)}\\{}=2\frac{(-1)^{n-1}/(n-1)!}{(-1)^{2n-1}/(2n-1)!}=(-1)^n\frac{(2n)!}{n!}.$$

Answer 2

We have $$\lim_{z\to n} \frac{\Gamma(-n-z+1)}{\Gamma(1-2z)} =2$$ I don't know how you "verified" the wrong limit since you don't share your math nor your code.

Answer 3

The issue comes from the value of n. If we plot the limit for various

n = {-2.00, -1.75, -1.50, ... 2.00}

from

z = [n-1,n+1]

we see that the limit flips back and forth between 1 and 2.

Table[
    Plot[Gamma[-z - n + 1]/Gamma[1 - 2*z],
    {z, n - 1, n + 1},
    PlotRange -> {{n - 1, n + 1}, {-5, 3}},
    PlotLabel -> Style["n = " <> ToString[n], 18],
    PlotStyle -> AbsoluteThickness[2],
    ImageSize -> Automatic -> 200,
    Epilog -> {
        {Black, Dashed, Line[{{n - 1, 2}, {n + 0.8, 2}}]},
        {Text[Style["1", Red, 18], {n + 0.9, 1}]},
        {Black, Dashed, Line[{{n - 1, 1}, {n + 0.8, 1}}]},
        {Text[Style["2", Red, 18], {n + 0.9, 2}]},
        {Red, Line[{{n, -5}, {n, 3}}]},
        {Text[Style["z=n", Red, 18], {n + 0.25, -4}]},
    }], {n, -2, 2, 0.25}]

(I'd attach an image, but my account is too new)

We see that for n = {0.5, 1.0, 1.5, 2.0,...} the limit is 2. Where all other values of n yield a limit of 1.

Notably, (and likely the cause of confusion on this post) if n=0, it will yield a limit of 1, where all other positive integers yield a limit of 2. So the result will change depending on if your assumption permits n to be zero or not.

You can also test this with the following limits:

In: n=.;

In: Limit[Gamma[-z - n + 1]/Gamma[1 - 2*z], z -> n,  Assumptions -> n \[Element] Integers && n >= 0]

Out: 1

In: Limit[Gamma[-z - n + 1]/Gamma[1 - 2*z], z -> n,  Assumptions -> n \[Element] Integers && n > 0]

Out: 2

There seems to be an issue with Mathematica. It should be giving a conditional answer for Assumptions -> n \[Element] Integers && n >= 0 depending on if n is zero, or greater than zero... but it is not.

Answer 4

For $n\ge 1$, using Legendre duplication formula, \begin{align*} \lim_{z\to n} \frac{\Gamma(1-n-z)}{\Gamma(1-2z)} &=\lim_{z\to n} \frac{\pi}{\sin(n\pi+z\pi)\Gamma(n+z)}\times \frac{\sin(2\pi z)\Gamma(2z)}{\pi}\\ &=(-1)^n\lim_{z\to n} \frac{\sin(2\pi z)}{\sin(\pi z)}\times \frac{\Gamma(2z)}{\Gamma(n+z)}\\ &=(-1)^n\lim_{z\to n} \frac{2\cos(2\pi z)}{\cos(\pi z)}\times \frac{\Gamma(2z)}{\Gamma(n+z)}\qquad ( L'Hospital's ~rule)\\ &=2\frac{(-1)^{2n}(-1)^n}{(-1)^n}\lim_{z\to n} \frac{\Gamma(2z)}{\Gamma(n+z)}\\ &=2\lim_{z\to n} \frac{\Gamma(2z)}{\Gamma(n+z)}=2 \end{align*} Hence \begin{align*} \lim_{z\to n} \frac{\Gamma(1-n-z)}{\Gamma(1-2z)} =\begin{cases} 1,\qquad if\quad n=0,\\ 2,\qquad if\quad n\geq 1, \end{cases} \end{align*}