Question

$1+\frac{1}{3!}+\frac{1}{5!}+\frac{1}{7!}+...$ to $\infty$ is equal to

• A. $\frac{1}{2}(e-e^{-1})$
• B. $\frac{1}{2}(e+e^{-1})$
• C. $\frac{1}{2}e$
• D. none of these

Answer 1

The correct option is A $\frac{1}{2}(e-e^{-1})$

$e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+...\infty$
$e^{-x}=1-x+\frac{x^2}{2!}-\frac{x^3}{3!}+...\infty$
$e^x-e^{-x}=2[x+\frac{x^3}{3!}+\frac{x^5}{5!}...\infty ]$
$\frac{1}{2}(e^x-e^{-x})=x+\frac{x^3}{3!}+\frac{x^5}{5!}...\infty$
Substituting x=1, we get
$\frac{e-e^{-1}}{2}=1+\frac{1}{3!}+\frac{1}{5!}...\infty$