Question

Find the sums of the infinite series
$1+\frac{1}{3}+\frac{1\cdot 3}{3\cdot 6}+\frac{1\cdot 3\cdot 5}{3\cdot 6\cdot 9}+....$

Answer 1

Let $x=1+\frac{1}{3}+\frac{1.3}{3.6}+\frac{\mathrm{1.3.5}}{\mathrm{3.6.9}}+\frac{\mathrm{1.3.5.7}}{\mathrm{3.6.9.12}}+....=1+1\left(\frac{1}{3}\right)+\frac{1.3}{1.2}{\left(\frac{1}{3}\right)}^2+\frac{\mathrm{1.3.5}}{\mathrm{1.2.3}}{\left(\frac{1}{3}\right)}^3+........$.......(i)

Observe that RHS is of the form

$1+p\left(\frac{k}{q}\right)+\frac{p(p+q)}{2!}{\left(\frac{k}{q}\right)}^2+\frac{p(p+q)(p+2q)}{3!}{\left(\frac{k}{q}\right)}^3+.......={(1-k)}^{-\frac{p}{q}}$.......(ii)

$($Binomial expansion for fractional index$)$
Comparing equation(i) and L.H.S of equation(ii), we get
$p=1,p+q=3$ and $\frac{k}{q}=\frac{1}{3}$
Substitute $p=1$ in $p+q=3$
$\Rightarrow q=3-1$
$\therefore q=2$
Substitute $q=2$ in $\frac{k}{q}=\frac{1}{3}$
$\Rightarrow k=\frac{1}{3}\times 2$
$\therefore k=\frac{2}{3}$

Hence, $p=1$ , $q=2$ , $k=\frac{2}{3}$
Substitute above values in R.H.S of equation(ii) to get the sum of the given series.

Therefore $x={(1-\frac{2}{3})}^{-\frac{1}{2}}$
$={\left(\frac{1}{3}\right)}^{-\frac{1}{2}}$
$={\left(\frac{3}{1}\right)}^{\frac{1}{2}}$ (Since, $(\frac{a}{b}{)}^{-m}=(\frac{b}{a}{)}^m$)
Hence, required sum is $x=\sqrt{3}$