Question

The sum of the series $\frac{1}{1.2}+\frac{1.3}{\mathrm{1.2.3.4}}+\frac{\mathrm{1.3.5}}{\mathrm{1.2.3.4.5.6}}+....$ is

• A. $e-1$
• B. $\sqrt{e}-1$
• C. $\sqrt{e}-2$
• D. $\sqrt{e}+e$

Answer 1

Answer: (B)

$\sqrt{e}-1$

The nth term of the given series is
$T_n=\frac{1\cdot 3\cdot 5\cdot 7\cdot ....\cdot (2n-1)}{1\cdot 2\cdot 3\cdot 4\cdot \mathrm{5...}\left(2n\right)}=\frac{1}{2\cdot 4\cdot 6\dots 2n}=\frac{1}{2^2(1\cdot 2\cdot 3\cdot \dots n)}=\frac{1}{2^{n}n!}$

$\therefore {\sum }_{n=1}^{\infty }{T}_n={\sum }_{n=1}^{\infty }\frac{{\left(1/2\right)}^n}{n!}={\sum }_{n=0}^{\infty }\frac{{\left(1/2\right)}^n}{n!}-1=e^{1/2}-1$