real analysis - Finding $lim_{n to infty}sum_{k=n}^{2n}frac{1}{k}$

Let $a_n$ be defined via
$a_n = \sum_{k=n}^{2n}\frac{1}{k}$. Compute, $\lim_{n\to\infty}a_n$.

I have an handwavy argument that since $\lim_{n\to\infty}\left(\sum_{k=1}^n \frac{1}{k} - \log(n) \right) = \gamma$, where $\gamma$ is the Euler-Mascheroni constant, I think that the limit above is $\log(2)$, but I am a bit confused. Can someone help me with this? Thanks!