Jtdylktuy

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 $t_{n}=left(frac{3}{4}right)^n quadfrac{5.7.9.dots dots (2n+3)}{4.6.8. dots dots (2n+2)}$

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 quadfrac{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.

Let
$$f(x)=sum_{n=1}^infty t_nx^n=sum_{n=1}^inftyfrac{5times7timescdots
times(2n+3)}{4times6timescdots
times(2n+2)}x^n.$$
Then
$$t_n=frac23frac1{(n+1)!}left(frac32right)left(frac52right)cdots
left(frac{2n+3}2right)=frac23u_{n+1}$$
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
$$f(x)=frac23sum_{n=1}^infty u_{n+1}x^n=frac23sum_{n=2}^infty
u_nx^{n-1}=frac2{3x}left(frac1{(1-x)^{3/2}}-1-frac{3x}2right)$$
Now insert $x=3/4$.