$\mathop {\lim }\limits_{n \to \infty } \frac{{{1^p} + {2^p} + {3^p} + ..... + {n^p}}}{{{n^{p + 1}}}} = $

  • A
    $\frac{1}{{p + 1}}$
  • B
    $\frac{1}{{1 - p}}$
  • C
    $\frac{1}{p} - \frac{1}{{p - 1}}$
  • D
    $\frac{1}{{p + 2}}$

Explore More

Similar Questions

$\lim _{n \rightarrow \infty}\left(\frac{n^{2}}{\left(n^{2}+1\right)(n+1)}+\frac{n^{2}}{\left(n^{2}+4\right)(n+2)}+\frac{n^{2}}{\left(n^{2}+9\right)(n+3)}+\ldots+\frac{n^{2}}{\left(n^{2}+n^{2}\right)(n+n)}\right)$ ની કિંમત શોધો.

$\mathop {\lim }\limits_{n \to \infty } \left[ {\frac{1}{n} + \frac{1}{{n + 1}} + \frac{1}{{n + 2}} + \dots + \frac{1}{{2n}}} \right] = $

Difficult
View Solution

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

Difficult
View Solution

$\mathop {\lim }\limits_{n \to \infty } \left( {\frac{{n + 1}}{{{n^2} + {1^2}}} + \frac{{n + 2}}{{{n^2} + {2^2}}} + \frac{{n + 3}}{{{n^2} + {3^2}}} + \dots + \frac{1}{n}} \right) = $

$\lim _{n \rightarrow \infty} \frac{\pi}{2 n}\left[\sin \frac{\pi}{2 n}+\sin \frac{2 \pi}{2 n}+\sin \frac{3 \pi}{2 n}+\ldots+\sin \frac{\pi}{2}\right]=$

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