The sum of the series $\frac{15}{16}+\frac{15}{16} \times \frac{21}{24}+\frac{15}{16} \times \frac{21}{24} \times \frac{27}{32}+\dots$

Suppose $ S=\frac{15}{16}+\frac{15}{16} \times \frac{21}{24}+\frac{15}{16} \times \frac{21}{24} \times \frac{27}{32}+\dots \dots$
Does it converge? If so find the sum.

What I attempted:- On inspection of the successive terms, it easy to deduce that the $ n^{th}$ term of the series is $ t_{n}=\left(\frac{3}{4}\right)^n \quad\frac{5.7.9.\dots \dots (2n+3)}{4.6.8. \dots \dots (2n+2)}$

Thus $ \frac{t_{n+1}}{t_n}=\frac{3}{4} \times \frac{2n+5}{2n+4}$ . As $ n \to \infty$ this ratio tends to $ \frac{3}{4}<1$ . Hence by Ratio test it turns out to be convergent.

A similar type of question has already been asked here. One of the commenters has provided a nice method to evaluate the sum of such series using recurrence relation and finally using the asymptotic form of Catalan Number.

To proceed exactly in the similar way, I wrote $ t_n$ as follows:- $ t_n=\frac{1}{3} \left(\frac{3}{8}\right)^n \quad\frac{\dots \dots (2n+3)}{(n+1)!}=\frac{1}{3} \left(\frac{3}{8}\right)^n \frac{n+2}{2^{n+2}} \binom{2n+4}{n+2}\approx \left(\frac{3}{4}\right)^{n-1}\frac{\sqrt{n+2}}{\sqrt{\pi}} \quad (\mbox{For large $ n$ })$ .

I have used the recurrence relation $ S_n=S_{n-1}+T_n$ , along with the initial condition $ S_1=\frac{15}{16}$ , in order to get a solution like this $ $ S_n=7.5+\frac{T_n^2}{T_n-T_{n-1}}$ $ .

I am getting trouble in evaluating the limit of the second term as $ n \to \infty$ .

I haven’t cross checked all the steps. Hope I would be pointed in case of any mistake.