In order to show that a series converges, I want to show that $\sum\ln(\frac{v_{n+1}}{v_n})$ Which led me to the following first part of the equation, but I didn't achieved to solve it so I looked in tha answer book for the second part wich I didn't understand... why does
\begin{align*} \ln(1+\frac{3}{n^2}+o(\frac{1}{n^2})) &=\frac{3}{n^2}+o(\frac{1}{n^2}) \end{align*}
Note that $$\ln(1+\frac{3}{n^2}+o(\frac{1}{n^2})) = \frac{3}{n^2}+o(\frac{1}{n^2}) \\ \iff \\ \frac{\ln(1+\frac{3}{n^2}+o(\frac{1}{n^2}))}{\frac{3}{n^2}} = \frac{\frac{3}{n^2}+o(\frac{1}{n^2})}{\frac{3}{n^2}} = 1 + \frac{o(\frac{1}{n^2})}{\frac{3}{n^2}} = 1+ 0 =1 $$ so you only need to prove $$\lim_{x \to 0} \frac{\ln(1+\frac{3}{n^2}+o(\frac{1}{n^2}))}{\frac{3}{n^2}} = 1$$ There are any workarounds here, but the simplest is probably using the Mac Laurin series, hence you get $$\lim_{x \to 0} \frac{\ln(1+\frac{3}{n^2}+o(\frac{1}{n^2}))}{\frac{3}{n^2}} = \frac{\frac{3}{n^2}+o(\frac{1}{n^2})}{\frac{3}{n^2}} = 1 + \frac{o(\frac{1}{n^2})}{\frac{3}{n^2}} = 1+ 0 =1$$
