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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: $$f(n)=2^{1+2n}B_{1/2}(n,n+2)$$ $$g(n)=4^nB_{1/2}(n+1,n)$$ $$ h(n)=n(n-1)/2 \left( \frac{f(n)}{n+1}-\frac{g(n)}{n-1}\right)$$ 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: $$B_z(a,b)=\int\limits_0^z u^{a-1}(1-u)^{b-1} \mathrm{d}u.$$

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  • $\begingroup$ $z \to B_z(a,b)$ 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 Kadelka Mar 5 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 Beenakker Mar 5 at 10:57
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Using the substitution $u=(1-s)/2$ in the relevant integrals $\int_0^{1/2} u^{a-1}(1-u)^{b-1} \,du$, 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 $n\to\infty$.

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    $\begingroup$ 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. $\endgroup$ – Carlo Beenakker Mar 5 at 16:08
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    $\begingroup$ @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. $\endgroup$ – Iosif Pinelis Mar 5 at 16:18
  • $\begingroup$ I meant the $h(n)=\frac{n}{4(n+1)}[\frac{(n-1)\Gamma(n)\sqrt{\pi}}{2\Gamma(n+3/2)}+3-\frac{1}{n}-\frac{2\Gamma(n)\sqrt{\pi}}{\Gamma(n+1/2)}]$. Probably it's same as the identity in the answer now. By the way, nice answer. $\endgroup$ – Alapan Das Mar 5 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. $\endgroup$ – Iosif Pinelis Mar 5 at 16:55
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The blue line is the numerical evaluation of $h(n)$, the gold line is

$$H(n)=\frac{3}{4}-\frac{3 \sqrt{\pi } n \Gamma (n)}{8 \Gamma \left(n+\frac{3}{2}\right)}\rightarrow \frac{3}{4}$$

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  • $\begingroup$ 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] $\endgroup$ – sajjad veeri Mar 5 at 8:20
  • $\begingroup$ that does not work, but I made an "educated guess", which seems to agree quite well numerically. $\endgroup$ – Carlo Beenakker Mar 5 at 10:05

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