Those fractions each tend to 0

Relevant wiki: Riemann Sums

alternatively,

$\large \displaystyle \lim_{n \to \infty} \displaystyle \sum_{i=1}^n \dfrac {1}{n+i}= \frac{1}{n} \sum_{i=1}^{n} \frac{1}{1+\frac{i}{n}}=\int_{1}^{2}\frac{dt}{t}=\ln2$

The summation can be rewritten as $H(2n) - H(n)$ , where $H(n)=\displaystyle \sum_{i=1}^{n} \frac{1}{i}$ . Thus, we wish to find

$L = \displaystyle \lim_{n \to \infty} H(2n) - H(n)$

Note that the Euler-Mascheroni constant is defined as

$\gamma = \displaystyle \lim_{n \to \infty} H(n)-\ln(n)$

Thus,

$L = \displaystyle \lim_{n \to \infty} \ln(2n)-\ln(n) = \ln(2) = 0.69$