$\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
    $\frac{1}{{p + 2}}$

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Similar Questions

$\lim _{n}$ ${\rightarrow \infty}\left[\left(1+\frac{1}{n^2}\right)\left(1+\frac{2^2}{n^2}\right) \ldots\left(1+\frac{n^2}{n^2}\right)\right]^{\frac{1}{n}}=$

योगफल की सीमा के रूप में $\int_{0}^{2} e^{x} dx$ का मूल्यांकन कीजिए।

$\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=0}^{2 n-1} \frac{n^{2}}{n^{2}+4 r^{2}}$ का मान है:

$a \in \mathbb{R}$ (सभी वास्तविक संख्याओं का समुच्चय) के लिए,$a \neq -1$,यदि $\lim_{n \to \infty} \frac{1^a + 2^a + \dots + n^a}{(n+1)^{a-1}[(na+1) + (na+2) + \dots + (na+n)]} = \frac{1}{60}$ है,तो $a$ का मान ज्ञात कीजिए:

यदि $U_{n}=\left(1+\frac{1^{2}}{n^{2}}\right)^{1}\left(1+\frac{2^{2}}{n^{2}}\right)^{2} \ldots\left(1+\frac{n^{2}}{n^{2}}\right)^{n}$ है,तो $\lim _{n \rightarrow \infty}\left(U_{n}\right)^{\frac{-4}{n^{2}}}$ का मान ज्ञात कीजिए:

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