Answer
$\Sigma_{n=0}^\infty \dfrac{x^{(2n+1)}}{(2n+1)!}$
Work Step by Step
Consider the Maclaurin Series for $\sin x$ and $e^x$ are as follows:
$ \sin x=\Sigma_{n=0}^\infty \dfrac{(-1)^n x^{2n+1}}{(2n+1)!}$
and
$e^x=\Sigma_{n=0}^\infty \dfrac{x^n}{n!}=1+x+\dfrac{x^2}{2!}+...$
Then,
$ \sinh x=\dfrac{e^x -e^{-x}}{2}$
$\implies \dfrac{1}{2}[(1+x+\dfrac{x^2}{2!}+...)-(1-x+\dfrac{x^2}{2!}+...)]=x+\dfrac{1}{3!}x^3+\dfrac{1}{5!} x^5=\Sigma_{n=0}^\infty \dfrac{x^{(2n+1)}}{(2n+1)!}$
