How to find a sum of $\sum_{n=1}^{\infty}(-1)^{n-1}n(n+1)x^n$?

Question

How to find a sum of $\sum_{n=1}^{\infty}(-1)^{n-1}n(n+1)x^n$?

I've got that the derivatives will look something like that $$(-1)^{n-k} \cdot n\dfrac{(n+1)!}{(n-k)!}x^{n-k}$$ where $k$ is a derivative number. The integral of starting sum looks like that $$(-1)^{n+1}nx^{n+1}$$ Can't get what to do next, I guess I should somehow compare it with the derivative of $\arctan$.

Answer 1

Hint. The derivative is the right tool but you have to differentiate twice! Note that for $|x|<1$, $$\sum_{n=1}^{\infty}(-1)^{n-1}n(n+1)x^n=x\frac{d^2}{dx^2}\left(\sum_{n=1}^{\infty}(-1)^{n-1}x^{n+1}\right)=x\frac{d^2}{dx^2}\left(\sum_{n=0}^{\infty}(-x)^{n}\right).$$

Answer 2

Hint:

Rewrite this series as $$x\sum_{n=1}^{\infty}(-1)^{n+1}n(n+1)x^{n-1}=x\biggl(\sum_{n=1}^{\infty}(-1)^{n+1}x^{n+1}\biggr)''=x\biggl(\sum_{n=0}^{\infty}(-1)^{n}x^{n}\biggr)''.$$