I am given the following series and asked to solve it. $\sum\limits_{n=1}^\infty \dfrac{(-2)^n}{(2n+1)!}$ I recognize that this series is somewhat similar to the Taylor series for $sinx$ which is $\sum\limits_{n=0}^\infty (-1)^n\dfrac{x^{2n+1}}{(2n+1)!}$.
However, I am not really able to relate these two series in order to solve them, especially since my series starts at 1 and once I rewrite the $sinx$ series to match that, I am completely lost.
Since $\sin(x)-x =\sum\limits_{n=1}^\infty (-1)^n\dfrac{x^{2n+1}}{(2n+1)!} $, $\dfrac{\sin(x)-x}{x} =\sum\limits_{n=1}^\infty (-1)^n\dfrac{x^{2n}}{(2n+1)!} =\sum\limits_{n=1}^\infty \dfrac{(-x^2)^{n}}{(2n+1)!} $.
Therefore, putting $\sqrt{x}$ for $x$, $\dfrac{\sin(\sqrt{x})-\sqrt{x}}{\sqrt{x}} =\sum\limits_{n=1}^\infty \dfrac{(-x)^{n}}{(2n+1)!} $.
And there you are.