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 $\frac{1}{1-x}=\sum x^n$?

  • $\begingroup$ This looks a lot like geometric series,try differentiating the geometric series $\endgroup$ – kingW3 Mar 15 '17 at 12:28
  • $\begingroup$ Hint: $x^{2n} = \left(x^2\right)^n$. $\endgroup$ – Math1000 Mar 15 '17 at 12:29
  • $\begingroup$ You hinting at the power series representation of $\sum [a_n](x-c)^n$ which gives $\sum [(-1)^{n-1}n]((x-0)^2)^n$? $\endgroup$ – themli Mar 15 '17 at 12:39

We have the alternating geometric series:


And we have its derivative:


Multiply both sides by $-x$ and let $x\mapsto x^2$ to finally get



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

  • $\begingroup$ One may likewise come up with this by following the finite form of my approach. (+1) and have a nice day! $\endgroup$ – Simply Beautiful Art Mar 15 '17 at 12:59
  • $\begingroup$ have a nice day too $\endgroup$ – Dr. Sonnhard Graubner Mar 15 '17 at 13:00

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