Why is $(1+\frac{3}{n})^{-1}=(1-\frac{3}{n}+\frac{9}{n^2}+o(\frac{1}{n^2}))$ and how to get around the Taylor expansion? 
Let be $(u_n)$ a real sequence such that $u_0>0$ and that $\forall n \in \mathbb{R}$:
$$\frac{u_{n+1}}{u_n}=\frac{n+1}{n+3}$$
Let be $(v_n)$ a real sequence such that $\forall n \in \mathbb{R}$:
$$v_n=n^2u_n$$
Let's determine the nature of $\sum\ln(\frac{v_{n+1}}{v_n})$

I did:
\begin{align*}
\frac{v_{n+1}}{v_n} &=\left(\frac{n+1}{n}\right)^2\frac{u_{n+1}}{u_n}\\
&=\left(1+\frac{1}{n}\right)^2\frac{n+1}{n+3}\\
&=\left(1+\frac{1}{n}\right)^3\left(1+\frac{3}{n}\right)^{-1}\\
\end{align*}
and there I was stuck. A friend of mine gave me this tip:
$$=\left(1+\frac{3}{n}+\frac{3}{n^2}+\frac{1}{n^3}\right)\left(1-\frac{3}{n}+\frac{9}{n^2}+o\left(\frac{1}{n^2}\right)\right)$$
But I don't understand this notation $o\left(\frac{1}{n^2}\right)$ I know that it means that its negligible. I checked on wikipedia that it describes the limiting behavior of a function when the argument tends towards a particular value or infinity
But still, I'm stuck:


*

*what does that mean, how do we end up to it?


I think it is related to the Taylor expansion of $\left(1+\frac{3}{n}\right)^{-1}$
because $(1+x)^\alpha=1+\alpha x+\frac{\alpha(\alpha -1)}{(2)!}x^2 + x^n\epsilon(x)$
then $\left(1+\frac{3}{n}\right)^{-1}=...$
but still I'm not very smart at usual Taylor developments so...


*is there a way to get around?

 A: 
I don't understand this notation $\displaystyle o\Big(\frac{1}{n^2}\Big)$

A general term $u_n$ is considered as being $\displaystyle o\Big(\frac{1}{n^2}\Big)$ if it is such that, as $n \to \infty$,
$$
\dfrac{u_n}{\frac{1}{n^2}}\to 0.
$$
One may recall that, by the Taylor series expansion, as $x \to 0$, one has
$$
\frac1{1+x}=1-x+x^2+o(x^2),\tag1
$$ here again, the notation $o(x^2)$ means any function $f$ satisfying, as $x \to 0$,
$$
\dfrac{f(x)}{x^2}\to 0.
$$
Then, to obtain an asymptotic expansion of $\displaystyle \left(1+\frac{3}{n}\right)^{-1} $, you may use $(1)$ with $x=\dfrac3n$, observing that as $n \to \infty$ we have $x \to 0$, giving
$$
\left(1+\frac{3}{n}\right)^{-1}=1-\frac{3}{n}+\frac{9}{n^2}+o\Big(\frac{1}{n^2}\Big)
$$ as given by your friend.
A: As jordan's comment notes, the sum is telescoping:  $$\begin{align*} \sum_{n=1}^N \log \frac{v_{n+1}}{v_n} &= \sum_{n=1}^N \log v_{n+1} - \log v_n \\ &= \log v_{N+1} + \sum_{n=2}^{N} \log v_n - \sum_{n=2}^N \log v_n - \log v_1 \\ &= \log v_{N+1} - \log v_1 \\ &= \log \frac{v_{N+1}}{v_1}. \end{align*}$$  Next, it is easy to see that $$\frac{u_{N+1}}{u_1} = \prod_{n=1}^N \frac{u_{n+1}}{u_n} = \prod_{n=1}^N \frac{n+1}{n+3} = \frac{(2)(3)}{(N+2)(N+3)}.$$  Therefore, the finite sum is $$\log \frac{(N+1)^2u_{N+1}}{u_1} = \log \frac{6(N+1)^2}{(N+2)(N+3)},$$ and the limit as $N \to \infty$ is simply $$\lim_{N \to \infty} \log \frac{6(N^2 + 2N + 1)}{N^2 + 5N + 6} = \log 6 + \lim_{N \to \infty} \log \frac{1 + 2N^{-1} + 1N^{-2}}{1 + 5N^{-1} + 6N^{-2}} = \log 6. $$
