Question

Find the sum of the following series $x+\frac{x^5}{\lfloor 5}+\frac{x^9}{\lfloor 9}+....$

Answer 1

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

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

$e^{ix}=1+ix+(-\frac{x^2}{2!})-\frac{{ix}^3}{3!}+\frac{x^4}{4!}+\frac{{ix}^5}{5!}+...$

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

$e^{ix}-e^{-ix}=2ix+2i\frac{x^5}{5!}+\frac{x^9}{9!}+.....$

$S=\frac{1}{2i}\{e^{ix}-e^{-ix}\}$