Question

The sum of series
$\frac{1}{1\cdot 2}+\frac{1\cdot 3}{1\cdot 2\cdot 3\cdot 4}+\frac{1\cdot 3\cdot 5}{1\cdot 2\cdot 3\cdot 4\cdot 5\cdot 6}+......\text{to}\infty$, is equal to

• A. $e-1$
• B. $e^{1/2}-1$
• C. $e^{\frac{1}{2}+e}$
• D. None of these

Answer 1

Answer: (B)

$e^{1/2}-1$

the $n$th term of the given series is
$T_n=\frac{1\cdot 3\cdot 5\cdot \mathrm{7....}(2n-1)}{1\cdot 2\cdot 3\cdot 4\cdot \mathrm{5....}\left(2n\right)}=\frac{1}{2n!}$
$\therefore {\sum }_{n=1}^{\infty }{T}_n={\sum }_{n=1}^{\infty }\frac{{\left(\frac{1}{2}\right)}^n}{n!}=\left({\sum }_{n=0}^{\infty }\frac{{\left(\frac{1}{2}\right)}^n}{n!}\right)-1$
$=e^{1/2}-1$