lim_(n→∞) ((1/(n^2 +1)) + (2/(n^2 +1)) + ... + ((n-1)/(n^2 +1))) = ?


Question and Answers Forum

All Questions Topic List
Algebra Questions
Previous in All Question Next in All Question
Previous in Algebra Next in Algebra
Question Number 157883 by HongKing last updated on 29/Oct/21

$\lim_{n \to \infty} (\frac{1}{n^{2} + 1} + \frac{2}{n^{2} + 1} + . . . + \frac{n - 1}{n^{2} + 1}) = ?$
	Answered by puissant last updated on 29/Oct/21
	$S=lim_(n→∞) ((1/(n^2 +1))+(2/(n^2 +1))+...+((n−1)/(n^2 +1))) = lim_(n→∞) (1/(n^2 +1)) Σ_(k=1) ^(n−1) k =lim_(n→∞) (((n−1)n)/(2(n^2 +1))) = (1/2)lim_(n→∞) ((n^2 (1−(1/n)))/(n^2 (1+(1/n^2 )))) =(1/2)lim_(n→∞) (({1−(1/n)})/({1+(1/n^2 )}))= (1/2).. ............Le puissant...........$
	$S = \lim_{n \to \infty} (\frac{1}{n^{2} + 1} + \frac{2}{n^{2} + 1} + . . . + \frac{n - 1}{n^{2} + 1})$ $= \lim_{n \to \infty} \frac{1}{n^{2} + 1} \underset{k = 1}{\sum^{n - 1}} k = \lim_{n \to \infty} \frac{(n - 1) n}{2 (n^{2} + 1)}$ $= \frac{1}{2} \lim_{n \to \infty} \frac{n^{2} (1 - \frac{1}{n})}{n^{2} (1 + \frac{1}{n^{2}})} = \frac{1}{2} \lim_{n \to \infty} \frac{{1 - \frac{1}{n}}}{{1 + \frac{1}{n^{2}}}} = \frac{1}{2} . .$ $. . . . . . . . . . . . L e p u i s s a n t . . . . . . . . . . .$
	Commented by Tawa11 last updated on 29/Oct/21

	$Great sir$
	Commented by HongKing last updated on 29/Oct/21

	$alot thanks sir$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum