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

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:

$$x^2\frac{1}{1-(-x^3)}$$

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

$$x^2\sum_{n=0}^{\infty}(-x^3)^n$$

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

$$x^2\sum_{n=0}^{\infty}(-1)^{n+1}x^{3n}$$

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

$$\sum_{n=0}^{\infty}(-1)^{n+1}x^{3n+2}$$

Is this correct?

• $$(-x^3)^n=(-1)^nx^{3n} \,.$$ – N. S. Dec 7 '13 at 3:05
• Inquisitor did a nice job of showing his effort here. – 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.
• Why $(-1)^n$ and not $(-1)^{n+1}$? Why should the first term of the sequence be positive instead of negative? – hax0r_n_code Dec 7 '13 at 3:07
• @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 $x^2 \displaystyle \sum_{n=0}^{\infty}(-1)^{n} x^{3n}$. – user17762 Dec 7 '13 at 3:09
• 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. – hax0r_n_code Dec 7 '13 at 3:11