First slide

Introduction to definite integration

Question

$\lim_{n \to \infty} (\frac{1^{m} + 2^{m} + 3^{m} + .... + n^{m}}{n^{m + 1}}) =____$

Moderate

A

$\frac{1}{m + 1}$
B

$\frac{1}{m + 2}$
C

$\frac{1}{m}$
D

$\frac{1}{m + 3}$

Solution

$\begin{array}{l} \lim_{n \to \infty} (\frac{1^{m} + 2^{m} + 3^{m} + \dots + n^{m}}{n^{m + 1}}) = \lim_{n \to \infty} \frac{1}{n} (\frac{1^{m}}{n^{m}} + \frac{2^{m}}{n^{m}} + \dots . + \frac{n^{m}}{n^{m}}) \\ = \lim_{n \to \infty} \frac{1}{n} \sum_{r = 1}^{n} {(\frac{r}{n})}^{m} = \int_{0}^{1} x^{m} d x = {(\frac{x^{m + 1}}{m + 1})}_{0}^{1} = \frac{1}{m + 1} - \frac{0}{m + 1} = \frac{1}{m + 1} \end{array}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App