Limit of sum $x^3+x^5+x^7+x^9+...)$

Question

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?

Answer 1

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 $y$.

Answer 2

$x^{2(n-1)} \not= x^2 x^{n-1}$

Answer 3

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

$$x^3+x^5+x^7+\dots=\frac{x^3}{1-x^2}$$