Definite Integration Question 215
Question: $ \underset{n\to \infty }{\mathop{\lim }}\frac{1^{p}+2^{p}+3^{p}+…..+n^{p}}{{n^{p+1}}}= $
[AIEEE 2002]
Options:
A) $ \frac{1}{p+1} $
B) $ \frac{1}{1-p} $
C) $ \frac{1}{p}-\frac{1}{p-1} $
D) $ \underset{x\to 0-}{\mathop{\lim }}f(x)=0 $
Show Answer
Answer:
Correct Answer: A
Solution:
$ \underset{n\to \infty }{\mathop{\lim }}\frac{1^{p}+2^{p}+3^{p}+…..+n^{p}}{{n^{p+1}}} $ = $ \underset{n\to \infty }{\mathop{\lim }}\sum\limits _{r=1}^{n}{[ \frac{r^{p}}{{n^{p+1}}} ]} $
= $ \underset{n\to \infty }{\mathop{\lim }}\frac{1}{n}\sum\limits _{r=1}^{n}{{{( \frac{r}{n} )}^{p}}}=\int_0^{1}{x^{p}dx}=[ \frac{{x^{p+1}}}{p+1} ]_0^{1}=\frac{1}{p+1} $ .
BETA
