Question

The sum of the series $\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}-...$ upto infinity is

• A. $e^{-2}$
• B. $e^{-1}$
• C. $e^{-1/2}$
• D. $e^{1/2}$

Answer 1

Answer: (B)

$e^{-x}=1-x+\frac{x^2}{2!}-\frac{x^3}{3!}+\frac{x^4}{4!}-...$
put $x=1$
$\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}...=e^{-1}.$