This question already has an answer here:
$$\lim_{n\rightarrow\infty}\sum_{k=1}^{n}\frac{1}{k+n} =\lim_{n\rightarrow\infty}\dfrac1n\sum_{k=1}^{n}\frac1{\dfrac kn+1}$$
Use $$\lim_{n \to \infty} \frac1n\sum_{r=1}^n f\left(\frac rn\right)=\int_0^1f(x)dx$$
$$\sum_{k=1}^n\frac1{k+n}~=~\sum_{k=n+1}^{2n}\frac1k~=~\sum_{k=1}^{2n}\frac1k~-~\sum_{k=1}^n\frac1k~=~H_{2n}-H_n~\simeq~\ln2n-\ln n~=~\ln\frac{2n}n$$ Can you take it from here, and evaluate what happens as $n\to\infty$ ? :-$)$