Mfrhtyj

Sum of series $displaystyle frac{15}{16}+frac{15}{16}cdotfrac{21}{24}+frac{15}{16}cdotfrac{21}{24}cdot frac{27}{32}+cdotscdots $

what i try

$displaystyle S =frac{15}{16}+frac{15cdot 21}{16cdot 24}+frac{15cdot 21cdot 27}{16cdot 24cdot 32}+cdots cdots $

i am trying to convert numerator and denomiantor terms into arithmetic progression

$displaystyle frac{9S}{8}=frac{9cdot 15}{8cdot 16}+frac{9cdot 15cdot 21}{8cdot 16cdot 24}+cdots cdots $

$displaystyle frac{9S}{8}+frac{9}{8}+1=1+frac{9}{8}+frac{9cdot 15}{8cdot 16}+frac{9cdot 15cdot 21}{8cdot 16cdot 24}+cdots cdots $

but it is divergent series

i did not know how i solve that infinite series

Help me how to solve

Based on the hint of lab bhattacharjee

$$S=frac{5}{2}left(frac{3}{8}right)+frac{5cdot7}{2cdot3}left(frac{3}{8}right)^2+frac{5cdot7cdot9}{2cdot3cdot4}left(frac{3}{8}right)^3+ cdots\
=frac{4}{3}cdotfrac{2}{3}sum_{n=2}^infty
frac{-frac{3}{2}left(-frac{3}{2}-1right)cdotsleft(-frac{3}{2}-n+1right)}{n!}left(-frac{3}{4}right)^n\
=frac{8}{9}left[left(1-frac{3}{4}right)^{-frac{3}{2}}-1-frac{-frac{3}{2}}{1!}left(-frac{3}{4}right)right]\=frac{8}{9}left[8-1-frac{9}{8}right]=frac{47}{9}.
$$

Generally for arbitrary $ainmathbb{R}_+$, $kinmathbb{Z}_+$, $|x|<1$:
$$
sum_{N=1}^infty prod_{n=1}^Nleft(1+frac{a-1}{k+n}right)x=
frac{k!}{a(a+1)cdots(a+k-1)x^k}sum_{N=k+1}^inftyfrac{a(a+1)cdots(a+N-1)}{N!}x^N\
=left[binom{a+k-1}{k}x^kright]^{-1}left[left(1-xright)^{-a}-sum_{N=0}^kbinom{a+N-1}{N}x^Nright].
$$

For your problem: $a=frac{3}{2}$, $k=1$, $x=frac{3}{4}$.

Note:
$$
binom{a}{k}:=frac{1}{k!}prod_{i=0}^{k-1} (a-i);quad
binom{a+k-1}{k}x^kequiv binom{-a}{k}(-x)^k.
$$

We have $S=frac{3}{4}sum_{n=2}^{infty}a_n$ with

$a_n=frac{5 cdot 7 cdot ... cdot (2n+1)}{4 cdot 6 cdot ... cdot (2n)}$. It is easy to see that $a_n ge 1$, hence $sum_{n=2}^{infty}a_n$ is divergent, thus $S= infty$.