shmm wrote: Find $ \lim_{n\to\infty}(\sum_{i=0}^{n}\frac{1}{n+i})$ Use definite integral to compute it

With $H_n=1+\frac{1}{2}+\dotsc+\frac{1}{n}$ and its well-known approximation this becomes $\lim_{n \to \infty} H_{2n}-H_{n-1}=\lim_{n \to \infty} \ln (2n) - \ln (n-1)=\lim_{n \to \infty} \ln \frac{2n}{n-1}=\ln 2 \approx 0.693$

$\sum_{i=0}^{n} \frac{1}{n+i}=\frac{1}{n}+\sum_{i=1}^{n} \frac{1}{n}\cdot \frac{1}{1+\frac{i}{n}}.$Thus $\lim_{n\to \infty} \left(\sum_{i=0}^{n} \frac{1}{n+i}\right)=\int_{0}^{1} \frac{dx}{1+x}$ $=\left[\ln (1+x)\right]_{0}^{1}=\boxed{\ln 2}\blacksquare$