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

Question Number 39443 by rahul 19 last updated on 06/Jul/18

\lim_{n \to \infty} [\frac{1}{n^{2} + 1} + \frac{2}{n^{2} + 2} + \frac{3}{n^{2} + 3} + \dots . + \frac{1}{n + 1}] = ?

Commented by tanmay.chaudhury50@gmail.com last updated on 06/Jul/18

v e r y g o o d \dots t h i s i s t h e w a y t o s o l v e i t \dots

Commented by math khazana by abdo last updated on 06/Jul/18

S_{n} = \sum_{k = 1}^{n} \frac{k}{n^{2} + k} w e h a v e 1 ⩽ k ⩽ n \Rightarrow

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

\Rightarrow \sum_{k = 1}^{n} \frac{k}{n^{2} + n} ⩽ \sum_{k = 1}^{n} \frac{k}{n^{2} + k} ⩽ \sum_{k = 1}^{n} \frac{k}{n^{2} + 1} \Rightarrow

\frac{1}{n^{2} + n} \frac{n^{2} + n}{2} ⩽ S_{n} ⩽ \frac{1}{n^{2} + 1} \frac{n^{2} + n}{2} \Rightarrow

\frac{1}{2} ⩽ S_{n} ⩽ \frac{n^{2} + n}{2 (n^{2} + 1)} \Rightarrow {l i m}_{n \to + \infty} S_{n} = \frac{1}{2} .

Commented by abdo mathsup 649 cc last updated on 07/Jul/18

t h a n k s .

Answered by ajfour last updated on 06/Jul/18

L e t n^{2} = N

L = \lim_{N \to \infty} N \underset{r = 1}{\sum^{\sqrt{N}}} \frac{(r / N) (\frac{1}{N})}{1 + \frac{r}{N}}

= \lim_{N \to \infty} N \int_{0}^{1 / \sqrt{N}} \frac{x d x}{1 + x}

= \lim_{N \to \infty} N [\frac{1}{\sqrt{N}} - \ln (1 + \frac{1}{\sqrt{N}})]

= \lim_{N \to \infty} N [\frac{1}{\sqrt{N}} - \frac{1}{\sqrt{N}} + \frac{1}{2 N} - \dots .]

\Rightarrow L = \frac{1}{2} .

Commented by rahul 19 last updated on 06/Jul/18

Ans. given is 1/2