Please, could you tell me if the following calculation is correct:
$$\sum\limits_{n = 0}^{\infty} \frac{\sin \frac{n \pi}{2}}{n!} x^n = \frac{0 x^0}{0!} + \frac{1 x^1}{1!} + \frac{0 x^2}{2!} + \frac{(-1) x^3}{3!}+\frac{0 x^4}{4!} + \frac{1 x^5}{5!} + \frac{0 x^6}{6!} + \frac{(-1) x^7}{7!} + \cdots =$$ $$= x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \frac{x^9}{9!} - \frac{x^{11}}{11!} + \frac{x^{13}}{13!} - \frac{x^{15}}{15!} + \cdots = $$ $$= (-1)^0 x^{2\cdot 0 + 1} + (-1)^1 \frac{x^{2\cdot 1 + 1}}{(2\cdot 1 + 1)!} + (-1)^2 \frac{x^{2\cdot 2 + 1}}{(2\cdot 2 + 1)!} + (-1)^3 \frac{x^{2\cdot 3 + 1}}{(2\cdot 3 + 1)!} + (-1)^4 \frac{x^{2\cdot 4 + 1}}{(2\cdot 4 + 1)!} + (-1)^5 \frac{x^{2\cdot 5 + 1}}{(2\cdot 5 + 1)!} + \cdots = $$ $$= \sum\limits_{n = 0}^{\infty} (-1)^n \frac{x^{2n + 1}}{(2n + 1)!}$$
This is from another question,
Development of the sine function into the power series
and I just want to be sure if I did everything correctly.