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. 
 A: Hint: The sum looks like the term by term derivative of $\sum (-1)^{n-1}\frac{x^{2n}}{(2n-1)!}$, that is, the derivative of $x\sum (-1)^{n-1}\frac{x^{2n-1}}{(2n-1)!}$. 
So we are basically looking at the derivative of something related to $x\sin x$.
To get the signs right, and take care of the fact that the sum does not start at $1$, I suggest writing down the first few terms of the expansion of $x\sin x$.  
Another way: We can get rid of the "awkward" $2n$ by writing $2n=(2n-1)+1$. The $2n-1$ partly cancels the $\frac{1}{(2n-1)!}$. 
Then the sum part looks like $\sum (-1)^{n-1}\frac{x^{2n-1}}{(2n-2)!}+\sum (-1)^{n-1} \frac{x^{2n-1}}{(2n-1)!}$. The first sum is closely connected to the series expansion of $x\cos x$. The second sum is also familiar. 
A: 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 $f(x)=-x +\sin x + x\cos x$.
