lim-x-1-n-2-pi-1-n-2-2pi-1-n-2-npi-

Question Number 120978 by ZiYangLee last updated on 04/Nov/20

\lim_{x \to \infty} (\frac{1}{n^{2} + π} + \frac{1}{n^{2} + 2 π} + \dots + \frac{1}{n^{2} + n π}) = ?

Commented by Dwaipayan Shikari last updated on 04/Nov/20

I f t h e q u e s t i o n b e c o m e s

\lim_{n \to \infty} (\frac{1}{n + π} + \frac{1}{n + 2 π} + \dots . + \frac{1}{n + n π})

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

= \int_{0}^{1} \frac{1}{1 + π x} d x = \frac{1}{π} \int_{0}^{π} \frac{d u}{1 + u} = \frac{1}{π} {[l o g (1 + u)]}_{0}^{π} = \frac{1}{π} l o g (1 + π)

Answered by Olaf last updated on 04/Nov/20

n^{2} + k π ⩾ n^{2}, k \in N^{*}

\frac{1}{n^{2} + k π} ⩽ \frac{1}{n^{2}}

0 ⩽ \underset{k = 1}{\sum^{n}} \frac{1}{n^{2} + k π} ⩽ \underset{k = 1}{\sum^{n}} \frac{1}{n^{2}} = n \times \frac{1}{n^{2}} = \frac{1}{n}

\lim_{n \to \infty} \frac{1}{n} = 0

\Rightarrow \lim_{n \to \infty} \underset{k = 1}{\sum^{n}} \frac{1}{n^{2} + k π} = 0

Facebook Tweet Pin