I am unable to follow one if the steps described in this question. My question is, can someone describe the part I do not understand?
The full question and solution is below :
Find the value for $x$ for which the series $U=\sum_{n=1}^\infty (2n-1)x^n$ converges.
$$(2n-1)(x^n)$$
Using the ratio test:
$$ \begin{align} R&=\lim_{n\to\infty}\left|\frac{U_{n+1}}{U_n}\right|=\lim_{n\to\infty}\frac{(2(n+1)-1)|x|^{n+1}}{(2n-1)|x|^n} \\ &=\lim_{n\to\infty} |x|\frac{2n+1}{2n-1}\\ &=\lim_{n\to\infty} |x|\frac{2+1/n}{2-1/n} \\ &\implies |x|<1 \end{align} $$
I understand that from the second line to the third line the fraction is divided by $n$ on both numerator and denominator.
However I do not follow the transition from the first line to the second, as the $-1$ term in the numerator seems to just disappear.
I would very much appreciate if someone could clarify the areas where I am unsure and how it works.
Looks like you're applying the ratio test to determine when the series converges. There are a couple of typos in your formulas and your typesetting. Here's how the reasoning should look: Write $u_n=(2n-1)x^n$. Then
$$ {u_{n+1}\over u_n} = {(2(n+1)-1 ) x^{n+1}\over (2n-1)x^n} = x\cdot {2n+1\over 2n-1} = x\cdot {2+\frac1n\over 2-\frac 1n}, $$ so $$\lim _{n\to\infty} \left|{u_{n+1}\over u_n}\right|=|x|. $$ (The transition from $(2(n+1)-1 ) $ to $2n+1$ is algebra.)