<s>[INST] 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.
 [/INST] *

*They are defining the terms of your sum. $u_1(x)=\frac{x^{1-1}}{2-1}=1$ and $u_2(x)=\frac{x^{2-1}}{4-1}=\frac{x}{3}$ and $u_3(x)=\frac{x^{3-1}}{6-1}=\frac{x^2}{5}$ and so on and so forth.


2.$|u_n|=\left|\frac{x^n}{2n+1}\right|$ but since $2n+1>0$ because we are dealing in positive integers then we can remove the absolute value around them resulting in $|u_n|=\frac{|x^n|}{2n+1}$ similar reasoning for $u_{n+1}$
3.basic exponent rules for division and a little bit of absolute value rules thrown in $\frac{|x^n|}{|x^{n-1}|}=\frac{|x|^n}{|x|^{n-1}}=|x|$ 
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