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!
Rewrite: $a_n = (H_{2n} - \log(2n))- (H_{n-1}- \log(n-1))+ \log\left(\dfrac{2n}{n-1}\right)$. Can you take it from here?
Almost in the same spirit as @DeepSea's answer $$a_n = \sum_{k=n}^{2n}\frac{1}{k}=H_{2 n}-H_{n-1}$$ Now, using the asymptotics of harmonic numbers $$H_p=\gamma +\log \left(p\right)+\frac{1}{2 p}-\frac{1}{12 p^2}+O\left(\frac{1}{p^3}\right)$$ Apply the above formula and continue with Taylor series to get $$a_n=\log (2)+\frac{3}{4 n}+\frac{1}{16 n^2}+O\left(\frac{1}{n^3}\right)$$ which is already "good" for rather small values of $n$. For example, $$a_{10}=\frac{178964263}{232792560}\approx 0.7687714$$ while the aboce formula would give $$a_{10}\approx\log (2)+\frac{121}{1600}\approx 0.7687722$$
