In trying to prove that the limit of a certain function approaches 1 as the positive integer parameter $n$ approaches infinity, I have ended up with the following intermediate expressions:
$$f(n)=2^{1+2n}B_{1/2}(n,n+2)$$
$$g(n)=4^nB_{1/2}(n+1,n)$$
$$ h(n)=n(n1)/2 \left( \frac{f(n)}{n+1}\frac{g(n)}{n1}\right)$$
Can somebody kindly help me with the evaluation of $ \lim_{n \to \infty} h(n)$?
If somebody could also plug it in Mathematica, I would be highly obliged. Thanks for any help in advance.
P.S.: In the above the notation $B_z(a,b)$ stands for the incomplete beta function defined by:
$$B_z(a,b)=\int\limits_0^z u^{a1}(1u)^{b1} \mathrm{d}u.$$

$\begingroup$ $z \to B_z(a,b)$ is not the incomplete beta function. You have to divide by $B(a,b) := B_1(a,b)$. $B(a,b)$ can be simply computed as a quotient of $\Gamma$functions. If I remember correctly its just $\frac{\Gamma(a+b)}{\Gamma(a)\Gamma(b)}$. $\endgroup$– Dieter KadelkaMar 5, 2021 at 10:44

$\begingroup$ @DieterKadelka  the formula in the OP uses the definition of the incomplete beta function on en.wikipedia.org/wiki/Beta_function#Incomplete_beta_function ; what you write is called the "regularized" incomplete beta function on Wikipedia. $\endgroup$– Carlo BeenakkerMar 5, 2021 at 10:57
3 Answers
Using the substitution $u=(1s)/2$ in the relevant integrals $\int_0^{1/2} u^{a1}(1u)^{b1} \,du$, then using the substitution $s^2=t$, and finally noting that $\Gamma(x+1/2)/\Gamma(x)\sim\sqrt x$ as $x\to\infty$, we have $$h(n)=\frac n{2(n+1)}\,\int_0^1 ds\,(1  s^2)^{n  1} ((n  1) s^2 + (3 n  1) s  2) \\ =\frac n{8(n+1)}\,{\left(6\frac2n\frac{3 \sqrt{\pi }\, \Gamma (n+2)}{n\Gamma \left(n+3/2\right)}\right)}\to\frac68$$ as $n\to\infty$.

1$\begingroup$ that cannot be quite correct, this expression diverges as $\sqrt n$ for large $n$; I think your factor $\Gamma(n+2)$ should be $n\Gamma(n)$ like in my surmise. $\endgroup$ Mar 5, 2021 at 16:08

1$\begingroup$ @CarloBeenakker : Thank you for your comment. There was a typo: the factor $n$ in the denominator $n\Gamma\left(n+3/2\right)$ was missing. this is now fixed. $\endgroup$ Mar 5, 2021 at 16:18

$\begingroup$ I meant the $h(n)=\frac{n}{4(n+1)}[\frac{(n1)\Gamma(n)\sqrt{\pi}}{2\Gamma(n+3/2)}+3\frac{1}{n}\frac{2\Gamma(n)\sqrt{\pi}}{\Gamma(n+1/2)}]$. Probably it's same as the identity in the answer now. By the way, nice answer. $\endgroup$ Mar 5, 2021 at 16:36

$\begingroup$ @AlapanDas : Thank you for your comment. Yes, your expression for $h(n)$ is the same as the one in this answer. $\endgroup$ Mar 5, 2021 at 16:55
The blue line is the numerical evaluation of $h(n)$, the gold line is
$$H(n)=\frac{3}{4}\frac{3 \sqrt{\pi } n \Gamma (n)}{8 \Gamma \left(n+\frac{3}{2}\right)}\rightarrow \frac{3}{4}$$

$\begingroup$ Mathematica can compute the limit as well .u may kindly copy f[n] = 2^(1 + 2 n) Beta[1/2, n, 2 + n] g[n] = 4^n Beta[1/2, 1 + n, n] f1[n] = f[n]/(n + 1) g1[n] = g[n]/(n  1) l[n] = n (n  1)/2 (f1[n]  g1[n]) Limit[l[n], n > Infinity] $\endgroup$ Mar 5, 2021 at 8:20

$\begingroup$ that does not work, but I made an "educated guess", which seems to agree quite well numerically. $\endgroup$ Mar 5, 2021 at 10:05
Python gives values upto $n=1000$. Values of $h(n)$ for $997 \leq n \leq 1000$ are $f(997)= 0.728174227325497 ,f(998)=0.7281857363386501 ,f(999)= 0.7281972275318511 ,f(1000)=0.7282087009506559$. Also, $f(900)=0.7269653136612485 $
The Code:
import matplotlib.pyplot as plt
import numpy as np
import math as m
def f(x):
f=0
g=0
h=0.001
for i in range(0,2000):
u=i*h
f=f+2*(u**(x1))*((1u/4)**(x+1))*h*(x1)
g=g+0.25*(u**x)*((1u/4)**(x1))*h*(x+1)
v=x/(x+1)*(fg)/2
return v
x=np.linspace(2, 1000, 1000)
F=np.vectorize(f)
plt.plot(x,F(x),'r')
plt.xlabel('n')
plt.ylabel('h(n)')
plt.show()

$\begingroup$ ,kindly share the python program as i am not able to access the file in colab because of permission issues $\endgroup$ Mar 5, 2021 at 9:54

$\begingroup$ scipy.sci.btdtr(a,b,x) in the module scipy computes the incomplete $B_x(a,b)$, I doubt if this is of any use for the limit due to floating point limits. $\endgroup$ Mar 5, 2021 at 10:07

$\begingroup$ @Dieter Kadelka thank you. I shall use this function. $\endgroup$ Mar 5, 2021 at 10:09