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 $\frac{1}{1-x}=\sum x^n$?
$\endgroup$ 32 Answers
$\begingroup$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
$\endgroup$ $\begingroup$$$\frac{x^2}{(1+x^2)^2}=\sum_{n=1}^\infty(-1)^{n-1}nx^{2n}$$
Hint: prove with induction that for the finite sum $$\sum_{n=1}^m(-1)^{n-1}nx^{2n}={\frac { \left( -{x}^{2} \right) ^{m+1} \left( \left( m+1 \right) {x} ^{2}-{x}^{2}+m+1 \right) }{ \left( {x}^{2}+1 \right) ^{2}}}+{\frac {{x }^{2}}{ \left( {x}^{2}+1 \right) ^{2}}} $$ is hold and compute then the Limit $$m$$ tends to infinity
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