Given the sequence $x_n=\frac{n}{n+1}$, find constants $a,b$ such that $x_n = 1 + O(n^{-a})$ and $x_n = 1+o(n^{-b})$
For $a$, I need
$$\lim_{n\to\infty}\frac{x_n-1}{n^{-a}}=\lim_{n\to\infty}\frac{\frac{n}{n+1}-1}{n^{-a}}=\lim_{n\to\infty}\frac{-1}{(n+1)n^{-a}}\in\Bbb{R}$$ but I'm not sure how to proceed.
For $b$, I need
$$\dots=\lim_{n\to\infty}\frac{-1}{(n+1)n^{-b}}=0$$ Here I think it's enough that $b \leq 0$, but I'm not sure how to justify that.
Long divide to find, $$ x_n=\frac{n}{n+1}=1-\frac{1}{n+1} $$ From which it is clear that $$ x_n=1+O(1/n) $$ since $$ x_n-1=\frac{1}{n+1}\leq \frac{1}{n} $$ for any $n$.
For the second part, you need some $b$ with $$ \lim_{n\to \infty}\frac{n^b}{n+1}=0 $$ but by inspection, you just need the order of $n$ in the numerator to be less than $1$. Indeed, $b=1-\epsilon$ for any $\epsilon>0$ will do.
Note that
$$\frac{n}{n+1}=\frac{1}{1+\frac1n}=\left(1+\frac1n\right)^{-1}=1-\frac1n+\frac1{n^2}...$$
Or without Taylor's expansion
$$\frac{n}{n+1}=\frac{n+1-1}{n+1}=1-\frac{1}{n+1}$$
Thus
$$x_n=1+O(n^{-1})=1+o(1)=1+o(n^{-b}) \quad b\in[0,1)$$
