I'm not sure if I solved this correctly, but here is the problem:

Find a power series for the following function $\frac{x^2}{1+x^3}$

And here is what I did:


Here is where I took some educated guesses as to how to setup the power series:


I then factored out the $-1$ like this:


Lastly I multiplied the $x^2$ through and got this:


Is this correct?

  • 2
    $\begingroup$ $$(-x^3)^n=(-1)^nx^{3n} \,.$$ $\endgroup$ – N. S. Dec 7 '13 at 3:05
  • 3
    $\begingroup$ Inquisitor did a nice job of showing his effort here. $\endgroup$ – ncmathsadist Dec 7 '13 at 3:30

You are almost right but for the exponent on $(-1)$. It should be $(-1)^{n}$ and not $(-1)^{n+1}$. Fix this and pat yourself on the back.

  • $\begingroup$ Why $(-1)^n$ and not $(-1)^{n+1}$? Why should the first term of the sequence be positive instead of negative? $\endgroup$ – hax0r_n_code Dec 7 '13 at 3:07
  • $\begingroup$ @inquisitor Note that you went from the step $x^2 \displaystyle \sum_{n=0}^{\infty}(-x^3)^n$ to $x^2 \displaystyle \sum_{n=0}^{\infty}(-1)^{n+1} x^{3n}$ instead of $x^2 \displaystyle \sum_{n=0}^{\infty}(-1)^{n} x^{3n}$. $\endgroup$ – user17762 Dec 7 '13 at 3:09
  • $\begingroup$ I see, I thought I needed to be $n+1$ to make the firm term negative since the index starts at $0$, but I think I see what you mean now. $\endgroup$ – hax0r_n_code Dec 7 '13 at 3:11

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.