Limit of sum $x^3+x^5+x^7+x^9+...)$ I am asked to give the limit of:
$$ x^3+x^5+x^7+x^9+... \quad x\in(-1,1)$$
So I do the following:
The sum of the first $n$ terms will be equal to:
$$x^3+x^5+x^7+...+x^{3+2(n-1)}$$
I factor out $x^3$, I get:
$$x^3(1+x^2+x^4+..+x^{2(n-1)})$$
I also factor out $x^2$, I get:
$$x^3 x^2(1/x^2+1+x^2+x^3+...+x^{n-1}=x^5(1/x^2+1+x^2+x^3+...+x^{n-1})$$
So I have:
$$x^5/x^2+x^5(1+x^2+x^3+...+x^{n-1})=x^3+\frac{x^5 (1-x^n)}{1-x}$$
So I take the limit when $n$ goes to infinity and I get:
$$x^3+\frac{x^5}{1-x}=\frac{x^3(1-x^2)}{1-x^2}+\frac{(1+x)x^5}{1-x^2}=\frac{x^3-x^5+x^5+x^6}{1-x^2}=\frac{x^3+x^6}{1-x^2}$$
But the right answer is $\frac{x^3}{1-x^2}$
Where did I go wrong?

The problem is that when you factor out $x^2$, the remaining powers should still differ by two, because multiplying powers is additive. What you really wanted to do is write $y=x^2$, and it becomes
$$\frac{x^3}{1-y}$$
and we can simply substitute $x^2$ for $y$. 
