$\mathop {\lim }\limits_{n \to \infty } \left[ {\frac{1}{n} + \frac{1}{{\sqrt {{n^2} + n} }} + \frac{1}{{\sqrt {{n^2} + 2n} }} + \dots + \frac{1}{{\sqrt {{n^2} + (n - 1)n} }}} \right]$ is equal to

  • A
    $2 + 2\sqrt{2}$
  • B
    $2\sqrt{2} - 2$
  • C
    $2\sqrt{2}$
  • D
    $2$

Explore More

Similar Questions

$\underset{n\to \infty }{\mathop{\lim }}\,\frac{{{1}^{99}}+{{2}^{99}}+{{3}^{99}}+......{{n}^{99}}}{{{n}^{100}}}=$

Difficult
View Solution

The value of $\lim _{n \rightarrow \infty} \sum_{k=1}^n \frac{n^3}{(n^2+k^2)(n^2+3k^2)}$ is:

$\lim _{n \rightarrow \infty}\left[\frac{n}{(n+1) \sqrt{2n+1}}+\frac{n}{(n+2) \sqrt{2(2n+2)}}+\frac{n}{(n+3) \sqrt{3(2n+3)}}+\ldots n \text{ terms}\right]=\int_0^1 f(x) d x$,then $f(x)=$

If $\mathop {\lim }\limits_{n \to \infty } \frac{{{1^a} + {2^a} + \dots + {n^a}}}{{{{\left( {n + 1} \right)}^{a - 1}}\left[ {\left( {na + 1} \right) + \dots + \left( {na + n} \right)} \right]}} = \frac{1}{{60}}$ for some positive real number $a$,then $a$ is equal to

$\lim _{n \rightarrow \infty}\left(\frac{1^2}{n^3+1^3}+\frac{2^2}{n^3+2^3}+\ldots+\frac{n^2}{n^3+n^3}\right)=$

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo