Question

State whether given statement is true or false

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

• A. True
• B. False

Answer 1

The correct option is A True

We know that

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

Put $x=1$, we get

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

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

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

Hence, this is the answer.