lim-n-1-n-1-1-n-2-1-n-3-1-n-n-

Question Number 95849 by bobhans last updated on 28/May/20

\lim_{n \to \infty} \frac{1}{n + 1} + \frac{1}{n + 2} + \frac{1}{n + 3} + \dots + \frac{1}{n + n} ? ?

Answered by john santu last updated on 28/May/20

\lim_{n \to \infty} \underset{k = 1}{\sum^{n}} (\frac{1}{n + k}) = \lim_{n \to \infty} \underset{k = 1}{\sum^{n}} \frac{1}{n} (\frac{1}{1 + \frac{k}{n}})

= \underset{1}{\int^{2}} \frac{d x}{x} = [\ln (x)]_{1}^{2} = \ln (2) .

Answered by mathmax by abdo last updated on 28/May/20

let U_{n} = \frac{1}{n + 1} + \frac{1}{n + 2} + \frac{1}{n + 3} + \dots . + \frac{1}{n + n} \Rightarrow U_{n} = \sum_{k = 1}^{n} \frac{1}{n + k}

= \frac{1}{n} \sum_{k = 1}^{n} \frac{1}{1 + \frac{k}{n}} \Rightarrow U_{n} is a Rieman sum \Rightarrow \lim_{n \to + \infty} U_{n} = \int_{0}^{1} \frac{dx}{1 + x}

= {[\ln (1 + x)]}_{0}^{1} = \ln (2)