If $a_n, b_n$ are positive sequence such that $a_n \to a>0$ then $\lim\inf_{n \to +\infty} a_nb_n=a\lim \inf_{n \to +\infty} b_n$

Question

If $\lim \inf_{n \to +\infty}b_n=+\infty$, it can be proved the result. Suppose $\lim \inf_{n \to +\infty}b_n=b$. Let $t<ab$, We have to show $t<a_{n}b_{n}<ab$ except for finite terms. We have $\frac{t}{a}<b$. Using definition of limit infimum it can be proved $\frac{t}{a}<b_n<b$ for all $n\geq N$ for some $N\in \mathbb{N}$. I don't know how to proceed further. Similarly if we take $ab<s$, it can be proved $b<b_n<\frac{s}{a}$ for infinitely many $n$. But I am stuck here.

Answer 1

There is some subsequence $(b_{n_k})_{k\in\Bbb N}$ of $(b_n)_{n\in\Bbb N}$ such that $\lim_{k\to\infty}b_{n_k}=b$. Therefore, $\lim_{k\to\infty}a_{n_k}b_{n_k}=ab$, and so $\liminf_n(a_nb_n)\leqslant ab$.

Suppose that $\liminf_n(a_nb_n)=s<ab$. Then there is a subsequence $(a_{n_k}b_{n_k})_{k\in\Bbb N}$ of $(a_nb_n)_{n\in\Bbb N}$ such that $\lim_{k\to\infty}a_{n_k}b_{n_k}=s$. But $\lim_{k\to\infty}a_{n_k}=a$. So,$$\lim_{k\to\infty}b_{n_k}=\frac sa<\frac{ab}a=b,$$which is impossible, since $b=\liminf_nb_n$.