0
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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
2
$\begingroup$

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

$$\frac{x^2}{(1+x^2)^2}=\sum_{n=1}^\infty(-1)^{n-1}nx^{2n}$$

$\endgroup$
1
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.