Question

Find the sum of the infinite series $1+\frac{1}{3}+\frac{1.3}{3.6}+\frac{\mathrm{1.3.5}}{\mathrm{3.6.9}}+....$

Answer 1

Sum of infinite series $1+\frac{1}{3}+\frac{1.3}{3.6}+\frac{\mathrm{1.3.5}}{\mathrm{3.6.9}}+...$

Let $S=1+\frac{1}{3(1!)}+\frac{1.3}{\left(3{)}^2\right(2!)}+\frac{\mathrm{1.3.5}}{\left(3{)}^3\right(3!)}+...$

using the binomial expansion for negative are national exponents i.e.

$(1-x{)}^{-\frac{m}{n}}=1+\frac{\left(\frac{-m}{n}\right)(-x)}{1!}+\frac{\left(\frac{-m}{n}\right)(\frac{-m}{n}-1)}{2!}(-x{)}^2+\frac{\left(\frac{-m}{n}\right)(\frac{-m}{n}-1)(\frac{-m}{n}-2)}{3!}(-x{)}^3+...$

$=1+\frac{m}{n}x+\frac{m(m+n)}{2!}{\left(\frac{x}{n}\right)}^2+\frac{m(m+n)(m+2n)}{3!}{\left(\frac{x}{n}\right)}^3+...$

comparing it with sum.

$S=1+\frac{1}{3}+\frac{1.3}{2^1}.\frac{1}{3^2}+\frac{\mathrm{1.3.5}}{3!}.\frac{1}{3^3}+...$

$\Rightarrow m=1,n=2$ & $\frac{x}{n}=\frac{1}{3}$

$\Rightarrow x=\frac{2}{3}$

$\Rightarrow S=(1-x{)}^{-m/n}={(1-\frac{2}{3})}^{-\frac{1}{2}}$

$={\left(\frac{1}{3}\right)}^{-1/2}=3^{1/2}=\sqrt{3}$