Knowing that :
prove that sequence definite by :
converge to a limit real of which they ask to calculate.
From the 'duoble inequality'...
$\displaystyle \int_{k}^{k+1} \frac{dx}{x} \le \frac{1}{k} \le \int_{k-1}^{k} \frac{dx}{x} $ (1)
... we derive...
$\displaystyle \sum_{k=n+1}^{2n} \int_{k}^{k+1} \frac{dx}{x} \le \sum_{k=n+1}^{2n} \frac{1}{k} \le \sum_{k=n+1}^{2n} \int_{k-1}^{k} \frac{dx}{x}$ (2)
... so that is...
$\displaystyle \int_{n+1}^{2n} \frac{dx}{x} \le u_{n} \le \int_{n}^{2n-1} \frac{dx}{x}$ (3)
... so that is...
$\displaystyle \ln \frac{2n}{n+1} \le u_{n} \le \ln \frac{2n-1}{n}$ (4)
From (4) is easy to conclude that is...
$\displaystyle \lim_{n \rightarrow \infty} u_{n} = \ln 2$ (5)
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$