I am confused with sequence where $o_{p}$ is involved. Assume we have the following sequence of random variables, defined on the same probability space: $$ x_{n} = \frac{M_{n}+a}{M_{n}(M_{n}+b)}, $$ where $a$ and $b$ are some positive constants and $M_{n} = o_{p}(\sqrt{n})$, with $o_{p}$ means small $o$ in probability, i.e. $\frac{M_{n}}{\sqrt{n}}$ converges to $0$ in probability.
In addition, we assume that $M_{n} > 0$ and for any integer $K$ we have $\mathbb{P}(\liminf\{M_{n}>K \}) =1$.
Can one make a conclusion about the limit of $\{x_{n}\}$?
We have using a power series expansion, if $M_n$ diverges, that $$\frac{M_{n}+a}{M_{n}(M_{n}+b)}=\frac{1}{M_n}+\frac{b-a}{M_n^2}+o(M_n^{-3}).$$
Say we have $f(x)=\frac{x+a}{x(x+b)}$ and say we are after a power series expansion. We then need a basis function, say we pick $\{x^{-n}\}$. Then we have $$\lim_{x\to \infty}xf(x)=1$$, which implies the first term in the power series expansion is $x^{-1}$. Now to the second term proceed as $$\lim_{x\to\infty} x^2\times(f(x)-\frac1x)=a-b$$ where $\frac1x$ is the first term. Therefore we have the second term to be $\frac{a-b}{x^2}$. The third term is hence $$\lim_{x\to\infty} x^3\times(f(x)-\frac1x-\frac{a-b}{x^2})=b(b-a)$$ and so on.