Let $$T_n ={\Big(\frac{n!}{1\cdot 3\cdot 5\cdot 7 \cdots (2n+1)}\Big)}^2,$$ compute $$\lim_{ n \to \infty} ( T_1+T_2+ \cdots +T_n)$$
I tried to find out a recurrence relation with the help of the given series but it did not help me. Provided that the limit exists, I tried to check the upper and lower bound for it but could not succeed.
Please help me in this regard. Thanks.
Notice $$T_n = \left(\frac{n!}{\prod_{\ell=1}^n 2\ell+1}\right)^2 = \left(\frac{(1)_n}{(\frac32)_n}\right)^2 4^{-n}$$ where $\displaystyle\;(\alpha)_n = \prod_{\ell=0}^{n-1}(\alpha+\ell)\;$ is the rising factorials.
Since $0 < (1)_n < (\frac32)_n$, we have $|T_n| < 4^{-n}$ and the series converges absolutely.
Aside from a constant, the sum at hand has the form of a generalized hypergeometric function. $$ \sum_{n=1}^\infty T_n = \sum_{n=1}^\infty \frac{(1)_n(1)_n(1)_n}{n!(\frac32)_n(\frac32)_n} 4^{-n} = {}_3F_2\left( 1, 1, 1 ; \frac32, \frac32; \frac14 \right) - 1$$
Throwing following command HypergeometricPFQ[{1,1,1},{3/2,3/2},1/4] - 1 to wolfram alpha, we can evaluate the series with high precision
$$\sum_{n=1}^\infty T_n \approx 0.132968794417890464087477244396881449530734610666632988747\ldots$$
Unluckily, there doesn't seem to be other simpler closed form for this number.
Note that $$\dfrac{\dfrac{(n+1)!}{(2n+3)!!}}{\dfrac{n!}{(2+1)!!}}=\frac{(n+1)!}{n!}\frac{(2n+1)!!}{(2n+3)!!}=\frac{n+1}{2n+3} \ \overset{n\to\infty}\longrightarrow \ \frac{1}{2}<1$$ and conclude with the ration test to prove convergence.
Note: $k!!$ denote the double factorial of $k$.
