Answer
$\Sigma_{n=0}^\infty \dfrac{x^{2n+1}}{(2n+1)!}$
Work Step by Step
Since, we know that the Maclaurin Series for $e^x$ is defined as:
$e^x=\Sigma_{n=0}^\infty \dfrac{x^n}{n!}=1+x+\dfrac{x^2}{2!}+...$
Now,
$ \sinh x=\dfrac{e^x -e^{-x}}{2}=\dfrac{1}{2}[(1+x+\dfrac{x^2}{2!}+...)-(1-x+\dfrac{x^2}{2!}+...)]$
or, $x+\dfrac{x^3}{3!}+\dfrac{x^5}{5!} =\Sigma_{n=0}^\infty \dfrac{x^{2n+1}}{(2n+1)!}$
