I'm trying to understand the solution of the following exercise:
Given $u_{n+1}=u_n-u_n^2=u_n(1-u_n)$ with $u_0=\alpha$ and $0<\alpha<1$, find the limit of $\dfrac{1}{u_{n+1}}-\dfrac{1}{u_n}$ and write an asymptotic expansion for $u_n$ as $$u_n=a+\dfrac{b}{n}+o\left( \dfrac{1}{n} \right)$$
Now, for the first part we can easily prove that $u_n \rightarrow 0$ for $n \rightarrow +\infty$, and so we have: $$\lim _{n \rightarrow +\infty} \dfrac{1}{u_{n+1}}-\dfrac{1}{u_n}=\lim _{n \rightarrow +\infty} \dfrac{1-1+u_n}{(1-u_n)u_n}=\lim _{n \rightarrow +\infty}\dfrac{1}{1-u_n}=1$$ The problem is the solution that the author gives for the second part: he says "because of the previous point, we have that: $$\dfrac{1}{u_{n+1}}=\dfrac{1}{u_n}+1+o(1) \qquad \qquad \qquad (1)$$ and so we obtain: $$\dfrac{1}{u_n}=n+o(n) \qquad \qquad \qquad (2)$$ and we can conclude that $u_n=\dfrac{1}{n}+o\left(\dfrac{1}{n}\right)$". I really do not understand the passage between $(1)$ and $(2)$. Thank you very much.