$1+\frac{1}{3}+\frac{1.3}{3.6}+\frac{1.3 .5}{3.6 .9}+\ldots$ to $\infty=$

Question

Aicharya · Accepted Answer

$1+\frac{1}{3}+\frac{1.3}{3.6}+\frac{1.3 .5}{3.6 .9}+\ldots$ to $\infty$ Let $S=1+\frac{1}{3(1)!}+\frac{1.3}{(3)^2(2!)}+\frac{1.3 .5}{(3)^3(3!)}+\ldots$ Comparing $(1-x)^{-\frac{m}{n}}=1+\frac{m}{n} x+\frac{m(m+n)}{2!}\left(\frac{x}{n}\right)^2+\frac{m(m+n)(m+2 n)}{3!}\left(\frac{x}{n}\right)^3+\ldots$ $m=1, n=2 \& x=\frac{2}{3} \Rightarrow S=(1-x)^{-\frac{m}{n}}=\left(1-\frac{2}{3}\right)^{-\frac{1}{2}}=\sqrt{3}$

$1+\frac{1}{3}+\frac{1.3}{3.6}+\frac{1.3 .5}{3.6 .9}+\ldots$ to $\infty=$

Solution