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

Question

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{1.3.5.7.9.\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.

Answer 1

Let $$f(x)=\sum_{n=1}^\infty t_nx^n=\sum_{n=1}^\infty\frac{5\times7\times\cdots \times(2n+3)}{4\times6\times\cdots \times(2n+2)}x^n.$$ Then $$t_n=\frac23\frac1{(n+1)!}\left(\frac32\right)\left(\frac52\right)\cdots \left(\frac{2n+3}2\right)=\frac23u_{n+1}$$ where $$u_n=\frac{(3/2)(5/2)\cdots((2n+1)/2)}{n!}.$$ Then, for $|x|<1$, $$\sum_{n=0}^\infty u_nx^n=\frac1{(1-x)^{3/2}}$$ by the binomial theorem. Then $$f(x)=\frac23\sum_{n=1}^\infty u_{n+1}x^n=\frac23\sum_{n=2}^\infty u_nx^{n-1}=\frac2{3x}\left(\frac1{(1-x)^{3/2}}-1-\frac{3x}2\right)$$ Now insert $x=3/4$.