mathop {lim }limits_{n to infty } frac{{{1^p} | vedclass

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(A) We are given the limit: $\mathop {\lim }\limits_{n 	o \infty } \frac{{{1^p} + {2^p} + {3^p} + ..... + {n^p}}}{{{n^{p + 1}}}}$This can be written

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$\frac{1}{{p + 1}}$

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(A) We are given the limit: $\mathop {\lim }\limits_{n 	o \infty } \frac{{{1^p} + {2^p} + {3^p} + ..... + {n^p}}}{{{n^{p + 1}}}}$This can be written in summation notation as: $\mathop {\lim }\limits_{n 	o \infty } \frac{1}{n} \sum\limits_{r = 1}^n {\left( {\frac{r}{n}} ight)^p}$Using the definition of a definite integral as the limit of a sum,$\mathop {\lim }\limits_{n 	o \infty } \frac{1}{n} \sum\limits_{r = 1}^n {f\left( {\frac{r}{n}} ight)} = \int_0^1 {f(x)dx}$,where $f(x) = x^p$.Thus,the integral becomes: $\int_0^1 {x^p dx}$Evaluating the integral: $\left[ {\frac{{{x^{p + 1}}}}{{p + 1}}} ight]_0^1 = \frac{1^{p+1}}{p+1} - \frac{0^{p+1}}{p+1} = \frac{1}{{p + 1}}$.

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

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