Finding the second derivative of an infinite series I'm asked to find the 2nd derivative of 
$$f(x)=-2x+\frac{2x^3}{3!}-\frac{2x^5}{5!}+\cdots+(-1)^{n+1}\frac{2x^{n+1}}{(2n+1)!}+\cdots=\sum \limits_{n=0}^{\infty} \frac{(-1)^{n+1}2x^{2n+1} }{(2n+1)!}$$
Applying the power rule to $f(x)$, I find I can simplify the coefficient, since $\frac{(2n+1)(2n)}{(2n-1)!}=\frac{1}{(2n-1)!}$. Also, because the first term of $f''(x)$ is positive, the $(-1)^{n+1}$ changes to $(-1)^{n}$, which leaves me with:
$$f''(x)=\sum \limits_{n=0}^{\infty} \frac{(-1)^n2x^{2n-1} }{(2n-1)!}$$
However, the answer given on the homework solution is $\sum \limits_{n=0}^\infty \frac{(-1)^n2x^{2n+1} }{(2n+1)!}$. I don't see how I can get from my answer to the given solution. 
 A: You did not take into account the constant term turning to zero under the derivative. If you look at the first term of your series you get $$\frac{2}{x (-1)!}$$ which should not be there. It might be helpful to write out the first few as you take derivatives. Your index should start at $n=1$.

On another note altogether, I believe the final series you wrote is simply $2\sin(x)$.
A: You have the right answer, but it is just written differently. One way to see this is by differentiating the series term by term instead. You will see the same series appear but with a minus sign. Or you can get there by relabeling your index, but you have to use a bit of care with the first terms. (This is one problem with your answer already: the n = 0 case doesn't make any sense.) 
A: $$f''(x)=2\sin(x)=-f(x)$$
You can see this when you note that
$$f(x)=\sum_{n=0}^{\infty}\frac{(-1)^{n+1}2x^{2n+1}}{(2n+1)!}=-2\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n+1}}{(2n+1)!}=-2\sin(x)$$
So,
$$\begin{array}{rcl}
f(x)&=&-2\sin(x)\\
f'(x)&=&-2\cos(x)\\
f''(x)&=&2\sin(x)\\
\end{array}$$
And
$$f''(x)=2\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n+1}}{(2n+1)!}$$
As noted in other answers, when you take the derivative of a power series, you need to eliminate the constant term. So, deriving two times forces the derivative to start at $n=2$. Then, just modify the index.
