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?

Note that
$$1+x^2+x^4+\dots =\sum_{n=0}^\infty x^{2n}=\sum_{n=0}^\infty (x^2)^n$$
is convergent for $x\in (-1,1)$. Thus, call 
$$S(x)=1+x^2+x^4+\dots$$
Then we have
$$S(x)=1+x^2\left(1+x^2+x^4+\dots\right)=1+x^2S(x)$$
Solving for $S(x)$ gives us
$$S(x)=\frac{1}{1-x^2}$$
(you could also get $S(x)$ by noting that it is a geometric series in $x^2$). Now, we have the relation that
$$x^3+x^5+x^7+\dots =x^3(1+x^2+x^4+\dots)=x^3 S(x)$$
Thus
$$x^3+x^5+x^7+\dots=\frac{x^3}{1-x^2}$$
