Suppose $x_n\in \left(1,1+\frac{\log n}{n}\right)$ for $n\in\mathbb{N}$. Does this imply $$ x_n\sim 1+\frac{\log n}{n}\text{as }n\to\infty? $$
My answer is "yes" since $x_n\in \left(1,1+\frac{\log n}{n}\right)$ implies $\lim_{n\to\infty}x_n=1$. Also $\lim_{n\to\infty}y_n=1$ for $y_n:=1+\frac{\log n}{n}$.
Because the limits of the sequences $(x_n)$ and $(1/y_n)$ exists (both are 1), we can compute $$ \lim_{n\to\infty}(x_n\cdot \frac{1}{y_n})=\lim_{n\to\infty}x_n\cdot\lim_{n\to\infty}(1/y_n)=1\cdot 1=1. (*) $$
But $(*)$ is what is meant with the notation $$ x_n\sim y_n \text{as }n\to\infty. $$