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

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} + \dots + \frac{n - 1}{n^{2} + 1}) = ?

Answered by puissant last updated on 29/Oct/21

S = \lim_{n \to \infty} (\frac{1}{n^{2} + 1} + \frac{2}{n^{2} + 1} + \dots + \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} . .

\dots \dots \dots \dots L e p u i s s a n t \dots \dots \dots . .

Commented by Tawa11 last updated on 29/Oct/21

Great sir

Commented by HongKing last updated on 29/Oct/21

alot thanks sir