How can we prove the following (similar) limits?
$$\sum_{k=1}^n \frac{1}{k} (\ln 2 - \frac{1}{n+2} - \frac{1}{n+3} - \cdots -\frac{1}{2n + 2}) \to 0. $$
$$\sum_{k=1}^n \frac{1}{k} (\ln 3 - \frac{1}{n+2} - \frac{1}{n+3} - \cdots -\frac{1}{3n + 3}) \to 0. $$
We have $$\sum_{k\leq n}\frac{1}{k}\left(\log\left(2\right)-\frac{1}{n+2}-\dots-\frac{1}{2n+2}\right)=H_{n}\left(\log\left(2\right)-\sum_{j=n+2}^{2n+2}\frac{1}{j}\right)=H_{n}\left(\log\left(2\right)-H_{2n+2}+H_{n+1}\right) $$ where $H_{n} $ is the $n $-th armonic number. Recalling that $$H_{n}=\log\left(n\right)+\gamma+O\left(\frac{1}{n}\right) $$ as $n\rightarrow\infty $ we have $$H_{n}\left(\log\left(2\right)-H_{2n+2}+H_{n+1}\right)=\left(\log\left(n\right)+\gamma+O\left(\frac{1}{n}\right)\right)\left(\log\left(2\right)-\log\left(2n+2\right)+\log\left(n+1\right)+O\left(\frac{1}{n}\right)\right)= $$ $$=\left(\log\left(n\right)+\gamma+O\left(\frac{1}{n}\right)\right)O\left(\frac{1}{n}\right)=O\left(\frac{\log\left(n\right)}{n}\right) $$ then your claim. The other case is similar. In fact you will get $$\left(\log\left(3\right)-\log\left(3n+3\right)+\log\left(n+1\right)+O\left(\frac{1}{n}\right)\right)=\left(\log\left(\frac{3n+3}{3n+3}\right)+O\left(\frac{1}{n}\right)\right)=O\left(\frac{1}{n}\right). $$