I have the power series $$\sum_{n=1}^{\infty}(-1)^{n-1}nx^{2n}=x^2-2x^4+3x^6+-…$$
In the previous exercise I found by using the ratio test, the interval of convergence to be $(-1,1)$.
I want to find a "simple" expression for the series above on its interval of convergence.
My first thought: it kind of looks like the Maclaurin representation of $\cos x$.
$$\cos(x)=\sum_{n=0}^{\infty}\frac{(-1)^{n}x^{2n}}{2n!}=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}$$
I could replace $\frac{1}{2n!}$ with $n$ (or can I?)
$$\sum_{n=0}^{\infty}(-1)^{n}nx^{2n}$$ Which is close but no cigar. Might be something obvious I'm missing here. Notice the starting index is different as well.
Alternatively I could do something with the geometric series $\frac{1}{1-x}=\sum x^n$?
Best Answer
We have the alternating geometric series:
$$\frac1{1+x}=\sum_{n=0}^\infty(-1)^nx^n$$
And we have its derivative:
$$\frac{-1}{(1+x)^2}=\sum_{n=1}^\infty(-1)^nnx^{n-1}$$
Multiply both sides by $-x$ and let $x\mapsto x^2$ to finally get