Answer
$$\tan^{-1} (\dfrac{2}{3})$$
Work Step by Step
The Taylor series for $\tan^{-1} x $ can be defined as: $\tan^{-1} x= x-\dfrac{x^3}{3}+\dfrac{ x^5}{5}-....; |x| \leq 1$
Consider the given series; we have:
$\\=\dfrac{2}{3} - \dfrac{2^3}{3^3 \cdot 3!} + \dfrac{2^5}{3^5 \cdot 5!}-....\\=\dfrac{2}{3}-( \dfrac{1}{3}) (\dfrac{2}{3})^3+( \dfrac{1}{5}) (\dfrac{2}{3})^5-....\\=\tan^{-1} (\dfrac{2}{3})$
