<s>[INST] Power series of trigonometric functions Problem statement: Determine those $x$, for which the power series is convergent and determine the sum.
$$f(x)=x+\sum_{n=2}^{\infty}(-1)^{n-1}2n\frac{x^{2n-1}}{(2n-1)!}$$.
Progress:
I have difficulty to handle the $2n$ inside the sum. I can clearly see, since $\sin x = \sum_{n=1}^{\infty}(-1)^{n-1}\frac{\large x^{2n-1}}{(2n-1)!}$, that $\sin x-x$ is involved. The term $2n$ is always even so I'm sure $\cos x$ will occur in the sum. 
 [/INST] This is similar to Andre's observation but more explicit. 
We have $f(x)=x+\sum_{n=2}^{\infty}(-1)^{n-1}2n\frac{x^{2n-1}}{(2n-1)!}$,
thus we can kill the $2n$ term by integrating under summation, giving us
$$
\begin{align*}
\int f(x) dx &= C + \frac{x^2}{2} + \sum_{n=2}^{\infty}(-1)^{n-1}\frac{x^{2n}}{(2n-1)!}
\\
&= C + \frac{x^2}{2} + \sum_{n=2}^{\infty}(-1)^{n-1}\frac{x^{2n}}{(2n-1)!}\cdot\frac{x}{x}
\\
&= C + \frac{x^2}{2} + x\sum_{n=2}^{\infty}(-1)^{n-1}\frac{x^{2n-1}}{(2n-1)!}
\\
&= C + \frac{x^2}{2} + x\left( \sum_{n=1}^{\infty}(-1)^{n-1}\frac{x^{2n-1}}{(2n-1)!}-x \right)
\\
&= C + \frac{x^2}{2} + x\left( \sin x-x \right)
\\
&= C - \frac{x^2}{2} + x\sin x.
\end{align*}
$$
Taking the derivative of the integral returns $f$, thus $f(x)=-x +\sin x + x\cos x$.
</s>