My attempt is as follows
$ \prod_{r=1}^n (2r-1)\cdot\prod_{r=1}^{n} (2r)= (2r)!$
$\prod_{r=1}^{n} (2r)= 2^n.r!$
The general term for the denominator or $2^{r-1}\frac{(r+1)!}{2}$
$T_r = \frac{\prod_{r=1}^n (2r-1)}{2^{r-2}(r+1)!} = \frac{(2r)!}{2^{2r-2}(r+1)!.r!} = \frac{\binom{2r}{r}}{(r+1)2^{2r-2}}$
How can I proceed with the summation?
Notice that you can write it as
$$1 + \frac{1\cdot3}{6} + \frac{1\cdot 3\cdot 5}{6\cdot 8} + \ldots = \ + \prod_{k = 1}^n \frac{2k+1}{2k+4}$$
The product gives you
$$\prod_{k = 1}^n \frac{2k+1}{2k+4} = \frac{4 \Gamma \left(n+\frac{3}{2}\right)}{\sqrt{\pi } \Gamma (n+3)}$$
Hence
$$1 + \frac{4 \Gamma \left(n+\frac{3}{2}\right)}{\sqrt{\pi } \Gamma (n+3)}$$
Now let's use this trick: multiply and divide by $\frac{\sqrt{\pi}}{2}$ the second term
$$\frac{\sqrt{\pi}}{2}\cdot \frac{2}{\sqrt{\pi}} \frac{4 \Gamma \left(n+\frac{3}{2}\right)}{\sqrt{\pi } \Gamma (n+3)}$$
But $\frac{\sqrt{\pi}}{2}$ is nothing but $\Gamma(3/2)$ hence
$$1 + \frac{8}{\pi} \underbrace{\frac{\Gamma(3/2) \Gamma \left(n+\frac{3}{2}\right)}{\Gamma (n+3)}}_{B\left(n + \frac{3}{2}, \frac{3}{2}\right)}$$
That is the celebrated Beta function.
For the records:
$$B(p, q) = \int_0^1 x^{p-1}(1-x)^{q-1}\ \text{d}x$$
In our case we need to evaluate the sum:
$$\sum_n B\left(n + \frac{3}{2}, \frac{3}{2}\right) = \sum_n \int_0^1 x^{n + 1/2} (1-x)^{1/2}\ \text{d}x = \int_0^1 x^{3/2} (1-x)^{-1/2}\ \text{d}x$$
The result of the integral is $3\pi/8$ hence
$$1 + \frac{8}{\pi} \cdot \frac{3\pi}{8} = 1+3 = 4$$