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. $e^{-\frac{1}{2}}$
• D. $e^{+\frac{1}{2}}$

Answer 1

The correct option is B $e^{-1}$

We know that $e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+...\infty$

Put $x=-1$

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

$\Rightarrow e^{-1}=\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}-\frac{1}{5!}+...\infty$