Question

The sum of the infinite series
$1+\frac{1}{3}+\frac{1.3}{3.6}+\frac{1.35}{\mathrm{3.6.9}}+\frac{\mathrm{1.3.5.7}}{\mathrm{3.6.9.12}}+$ ... is equal to

• A. $\sqrt{2}$
• B. $\sqrt{3}$
• C. $\sqrt{\frac{3}{2}}$
• D. $\sqrt{\frac{1}{3}}$

Answer 1

Answer: (B)

Given series, $1+\frac{1}{3}+\frac{1\cdot 3}{3\cdot 6}+\frac{1\cdot 3\cdot 5}{3\cdot 6\cdot 9}+\frac{1\cdot 3\cdot 5\cdot 7}{3\cdot 6\cdot 9\cdot 12}+\dots \infty$
$=1+\frac{1}{3}+\frac{\frac{1}{2}\cdot \frac{3}{2}}{1\cdot 2}{\left(\frac{2}{3}\right)}^2+\frac{\frac{1}{2}\cdot \frac{3}{2}\cdot \frac{5}{2}}{1\cdot 2\cdot 3}{\left(\frac{2}{3}\right)}^3\dots \infty$
$=1+\frac{1}{2}\cdot \frac{2}{3}+\frac{\frac{1}{2}\left(\frac{1}{2}+1\right)}{2!}{\left(\frac{2}{3}\right)}^2+\frac{\frac{1}{2}\left(\frac{1}{2}+1\right)\left(\frac{1}{2}+2\right)}{3!}$
${\left(\frac{2}{3}\right)}^3\dots \infty$
$={\left(1-\frac{2}{3}\right)}^{-1/2}=3^{1/2}=\sqrt{3}$