A) \[\frac{1}{p+1}\]
B) \[\frac{1}{1-p}\]
C) \[\frac{1}{p}-\frac{1}{p-1}\]
D) \[\underset{x\to 0-}{\mathop{\lim }}\,f(x)=0\]
Correct Answer: A
Solution :
\[\underset{n\to \infty }{\mathop{\lim }}\,\frac{{{1}^{p}}+{{2}^{p}}+{{3}^{p}}+.....+{{n}^{p}}}{{{n}^{p+1}}}\]= \[\underset{n\to \infty }{\mathop{\lim }}\,\sum\limits_{r=1}^{n}{\left[ \frac{{{r}^{p}}}{{{n}^{p+1}}} \right]}\] = \[\underset{n\to \infty }{\mathop{\lim }}\,\frac{1}{n}\sum\limits_{r=1}^{n}{{{\left( \frac{r}{n} \right)}^{p}}}=\int_{0}^{1}{{{x}^{p}}dx}=\left[ \frac{{{x}^{p+1}}}{p+1} \right]_{0}^{1}=\frac{1}{p+1}\].
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