For what values of x does the series $1+\frac{x}{3}+\frac{x^2}{5}+\frac{x^3}{7}+\cdot\cdot\cdot$ converge? For what values of x does the series $1+\frac{x}{3}+\frac{x^2}{5}+\frac{x^3}{7}+\cdot\cdot\cdot$ converge?
The solution states:
The general term is of the form $u_n(x)=\frac{x^{n-1}}{(2n-1)}$, and hence
$$\frac{|u_{n+1}|}{|u_n|}=\frac{|x^n|}{(2n+1)}\cdot\frac{(2n-1)}{|x^{n-1}|}$$
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$$=\frac{(2n-1)}{(2n+1)}\cdot\frac{|x^n|}{|x^{n-1}|}$$
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$$=\frac{(2n-1)}{(2n+1)}|x|$$
clearly
$$\lim\limits_{n\rightarrow\infty}\frac{|u_{n+1}|}{|u_n|}=|x|$$
My question is:  


*

*How do you get from $1+\frac{x}{3}+\frac{x^2}{5}+\frac{x^3}{7}+\cdot\cdot\cdot$ to $u_n(x)=\frac{x^{n-1}}{(2n-1)}$?

*Why do the absolute restriction only apply to the $|x^n|$ and $|x^{n-1}|$ in the next line and not to the rest of the equation?

*and lastly, We go from $=\frac{(2n-1)}{(2n+1)}\cdot\frac{|x^n|}{|x^{n-1}|}$ to $=\frac{(2n-1)}{(2n+1)}|x|$ I.E. how does $\frac{|x^n|}{|x^{n-1}|}=|x|$ in the next line? (I understand that once you apply the limits, that $\frac{(2n-1)}{(2n+1)} = 1$)


Edits
I have added a line to the equation that was not there earlier and clarified my last question.

$$\frac{x^0}{1}+\frac{x^1}{3}+\frac{x^2}{5}+\frac{x^3}{7}+\cdot\cdot\cdot$$ 
You can see the pattern,  in the numerator It's:
$$x^0, x^1, x^2, x^3, \dots, x^{n-1}$$
And in the denominator, it's the odd numbers:
$$1, 3, 5, 7, 9, \dots, 2n-1$$
Therefore, it is:
$$u_n(x)=\frac{x^{n-1}}{(2n-1)}$$
While all odd numbers are positive ($n>0$), you can take them out of the absolute and $2n+1$ produce the same series of number as $2n-1$ which is odd numbers:
$$\frac{|u_{n+1}|}{|u_n|}=\frac{|x^n|}{\color{red}{(2n+1)}}\cdot\frac{\color{red}{(2n-1)}}{|x^{n-1}|} = \frac{|x^n|}{|x^{n-1}|}=|\frac{x^n}{x^{n-1}}|=|x|$$
Hence:
$$\lim\limits_{n\rightarrow\infty}\frac{|u_{n+1}|}{|u_n|}=|x|$$
