Consider the summation below: $$\sum_{k=1}^\infty(2k+1)x^{2k}$$ The problem asks to find what the mentioned summation is equal to. The solution provided in the book starts like the following: $$x^3+x^5+x^7+...=\frac{x^3}{1-x^2}$$ and then takes the derivative of both sides and uses that to find the solution.
My question is shouldn't $x^3+x^5+x^7+...=\frac{x^3(1-x^{n+1})}{1-x^2}$?
How does the book assume that $x^3+x^5+x^7+...$ equals $\frac{x^3}{1-x^2}$ when we're not told whether $\lvert x\rvert<1$ or not?
Let's make some order.
As @Crostul and @cosmo5 pointed out in the comments we clearly have to suppose $|x| < 1$ otherways the series diverges. Then using only some known properties of geometric series: $$ x^3 + x^5 + \dots + x^{3+2n} = \sum_{n=0}^N x^{3 + 2j} = x^3 \cdot \sum_{n=0}^N (x^2)^j = x^3 \cdot \frac{1-(x^2)^{N+1}}{1-x^2} = \frac{x^{3}(1-x^{2(N+1)})}{1-x^2}$$
Since we have supposed $|x| < 1$ we can state $$ \sum_{n=0}^\infty x^{3 + 2n} = \frac{x^{3}}{1-x^2} \qquad \qquad (1)$$
Now you can take what suggest your book [see somewhere why you can do the $\star$ passage!]
Differentiating the left part in $(1)$ we get: $$ \frac{d}{dx}\sum_{n=0}^\infty x^{3 + 2n} \stackrel{\star}{=} \sum_{n=0}^\infty \frac{d}{dx} \left(x^{3 + 2n} \right)= \sum_{n=0}^\infty (3+2n)x^{2n+2} = 3x^2 + \sum_{n=1}^\infty (3+2n)x^{2n+2}$$
Differentiating the right part in $(1)$ we get: $$ \frac{d}{dx} \left( \frac{x^{3}}{1-x^2} \right)= \frac{x^2(3 - x^2)}{(1-x^2)^2}$$
Dividing both results for $x^2$ we get: $$ 3 + \sum_{n=1}^\infty (3+2n)x^{2n} = \frac{(3 - x^2)}{(1-x^2)^2}$$
Split $$\sum_{n=1}^\infty (3+2n)x^{2n} = \sum_{n=1}^\infty (1+2n)x^{2n} + 2\sum_{n=1}^\infty x^{2n}$$
and you get: $$\begin{align}\sum_{n=1}^\infty (1+2n)x^{2n} & = \frac{(3 - x^2)}{(1-x^2)^2} -3 - 2\sum_{n=1}^\infty x^{2n} \\[10pt] & = \frac{(3 - x^2)}{(1-x^2)^2} - 3- \frac{2x^2}{1-x^2} \\[10pt] & = \frac{3x^2 - x^4}{(1-x^2)^2} \end{align}$$