Black Holes and the Expanding Universe

The radius is given by the expression
$r(t) = \bigg(\sum_{n=0}^t \frac{(-1)^n}{2n+1} \binom{t}{n}\bigg)^{-1}$

To evaluate the sum involving the combinatoric coefficients, consider the expansion,

$(1-x^2)^n = \sum_{r=0}^n (-1)^n \binom{n}{r} x^{2r}$ Integrating on both the sides with limits 0 to 1, we get,

$\int\limits_0^1 (1-x^2)^n dx = \sum_{r=0}^n (-1)^r \binom{n}{r} \int\limits_0^1x^{2r} dx$

$\Rightarrow \int\limits_0^1 (1-x^2)^n dx = \sum_{r=0}^n \frac{(-1)^r}{2r+1} \binom{n}{r}$

The RHS of the above equation is the required sum. Let the LHS be $I_{n}$ .Therefore,

$I_{n} = \int\limits_0^1 (1-x^2)^n dx$

$\Rightarrow I_{n} = \left.x(1-x^2)^n\right|_0^1 - \int\limits_0^1 (-2n)x^2 (1-x^2)^{n-1} dx$

$\Rightarrow I_{n} = 2n \int\limits_0^1 (x^2 -1 +1)(1-x^2)^{n-1} dx$ On rearranging slightly, we obtain,
$\frac{I_{n}}{2n} = - I_{n} + I_{n-1}$
$\Rightarrow \frac{I_{n}}{I_{n-1}} = \frac{2n}{2n+1}$
Now, Clearly $I_{0} = 1$ . To find $I_{n}$ consider the product, $\prod_{r=1}^n \frac{I_{r}}{I_{r-1}} = \prod_{r=1}^n \frac{2r}{2r+1}$

LHS of the above expression evaluates to $\frac{I_{n}}{I_{0}} = I_{n}$ . And, $r(t) = \frac{1}{I_{t}}$ .Thus,
$r(t) = \prod_{r=1}^n \frac{2r+1}{2r}$
$\Rightarrow r(t) = \frac{3}{2} \times \frac{5}{4} \times \ldots \times \frac{2t+1}{2t}$
$\Rightarrow r(t) = \frac{1\times 2 \times 3 \times 4 \times 5 \times \ldots \times 2t \times (2t+1)}{(2\times 4 \times \ldots \times 2t)^2}$
$\Rightarrow r(t) = \frac{ (2t)!(2t+1) }{ 4^t (t!)(t!) }$
$\Rightarrow r(t) = \frac{2t+1}{4^t} \binom{2t}{t}$

Now, number of black holes = $2t+1$ . Therefore, the number of black holes per unit area can be given as,

$\rho = \lim_{t \to \infty}\frac{2t+1}{ 4\pi (r(t))^2}$ $\rho = \lim_{t \to \infty}\frac{2t+1}{ 4\pi \bigg(\frac{2t+1}{4^t} \binom{2t}{t} \bigg)^2}$

Using Stirling approximation , we can write, $n! = \sqrt{2\pi n}\bigg(\frac{n}{e}\bigg)^n$ and,
$\binom{2n}{n} = \frac{(2n)!}{(n!)(n!)}$
Therefore,
$\binom{2t}{t} = \frac{4^t}{\sqrt{\pi t}}$
$\Rightarrow \rho = \lim_{t \to \infty}\frac{2t+1}{ 4\pi \bigg(\frac{2t+1}{4^t} \times \frac{4^t}{\sqrt{\pi t}} \bigg)^2}$
On rearranging, we obtain,
$\Rightarrow \rho = \lim_{t \to \infty} \frac{(2t+1)\pi t }{4\pi (2t+1)^2}$
Taking limits, we obtain, $\boxed{ \rho = \frac{1}{8}}$