Answer
$1-\dfrac{(x^{5/3})^2}{2!}+\dfrac{(x^{5/3})^4}{4!}-...=\Sigma_{n=0}^\infty (-1)^n\dfrac{ (x)^{(10n/3)}}{(2n)!}$
Work Step by Step
Consider the Taylor Series for $\cos x$ as follows:
$ \cos x=\Sigma_{n=0}^\infty \dfrac{(-1)^n x^{2n}}{(2n)!}=1-\dfrac{x^2}{2!}+\dfrac{x^4}{4!}-...$
Replace $(x^{5/3})$ with $x$ in the above series.
$\cos (x^{5/3})=\Sigma_{n=0}^\infty \dfrac{(-1)^n (x^{5/3})^{2n}}{(2n)!}$
This implies that $1-\dfrac{(x^{5/3})^2}{2!}+\dfrac{(x^{5/3})^4}{4!}-...=\Sigma_{n=0}^\infty (-1)^n\dfrac{ (x)^{(10n/3)}}{(2n)!}$
