TITLE: Find sum to infinite terms of the series $S=\frac{4}{5}+\frac{4.7}{5.8}+\frac{4.7.10}{5.8.11}+\cdots$
QUESTION [5 upvotes]: Find sum to infinity terms of the series $$S=\frac{4}{5}+\frac{4.7}{5.8}+\frac{4.7.10}{5.8.11}+\cdots$$
My Try:
we have $$1+S=1+\frac{4}{5}+\frac{4.7}{5.8}+\frac{4.7.10}{5.8.11}+\cdots$$
now $$(1+x)^n=1+nx+\frac{n(n-1)x^2}{2}+\cdots $$  So comparing we get
$$nx=\frac{4}{5}$$ and
$$\frac{n(n-1)x^2}{2}=\frac{7}{10}$$
solving for $x$ and $n$ we get
$$n=\frac{-16}{19}$$ and $$x=\frac{-19}{20}$$
Hence
$$1+S=(1+x)^n=20^{\frac{16}{19}}$$
$$S=20^{\frac{16}{19}}-1$$
Is this alright?

REPLY [4 votes]: $$S=\sum_{n\geq 1}\prod_{k=1}^{n}\frac{3k+1}{3k+2}=\sum_{n\geq 1}\frac{2^{2/3} \,\Gamma\left(\tfrac{5}{6}\right)\, \Gamma\left(\tfrac{4}{3}+n\right)}{\sqrt{\pi}\,\Gamma\left(\tfrac{5}{3}+n\right)} $$
is divergent by Gautschi's inequality and the p-test: the main term of such series behaves like $\frac{C}{n^{1/3}}$.