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

WBJEEWBJEE 2012Sequences and Series

Solution:

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