Example of divergent improper integral

Let's assume that $$f:[1,\infty) \to [0,\infty)$$.

Can someone provide an example of a function where the improper integral $$\int\limits_1^{\infty} f(x)dx$$ doesn't exist because we can find two sequences $$\left(a_n\right)_{n\in\mathbb {N}}$$ and $$\left(b_n\right)_{n\in\mathbb {N}}$$ with $$\lim\limits_{n\to\infty}a_n=\lim\limits_{n\to\infty}b_n=\infty$$ such that $$\lim\limits_{n\to\infty}\int\limits_1^{a_n} f(x)dx<\infty$$, $$\lim\limits_{n\to\infty}\int\limits_1^{b_n} f(x)dx<\infty$$ and $$\lim\limits_{n\to\infty}\int\limits_1^{a_n} f(x)dx \neq \lim\limits_{n\to\infty}\int\limits_1^{b_n} f(x)dx$$?

I don't think such a case is possible, indeed $$g : x \rightarrow \int_1^x f(t)dt$$ is a monotonous function and as such admits a limit as $$x \to \infty$$. This limit can be either finite or $$+\infty$$ but in either case for any sequence $$x_n$$ that goes to $$+\infty$$, we have that $$lim_{n\to\infty}g(x_n) \to lim_{x\to\infty}g(x)$$.
• Supposed we don't know that $g$ admits a limit (finite or infinite). Why is it not possible that $\int\limits_1^{x_n}f(t)dt$ jumps between two finite values as $x_n \to \infty$ where $x_n$ is not monotone? Commented Oct 16, 2020 at 18:02
• Basically even if $x_n$ oscillates while going up to $\infty$, it still goes up to $\infty$ meaning that for any $M>0$, there will be a $n_0$ such that $x_n>M$ as long as $n>n_0$. Basically, at some point you can't go back down even if you oscillate. Commented Oct 19, 2020 at 11:14