$\lim_{n \to \infty} \frac{\sqrt{1} + \sqrt{2} + \dots + \sqrt{n}}{n^{\frac{3}{2}}} =$

  • A
    $0$
  • B
    $\frac{2}{3}$
  • C
    $1$
  • D
    $\frac{3}{2}$

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Similar Questions

$\lim _{n \rightarrow \infty} \frac{1}{n} \left\{ \sec ^{2} \frac{\pi}{4 n} + \sec ^{2} \frac{2 \pi}{4 n} + \ldots + \sec ^{2} \frac{n \pi}{4 n} \right\}$ ની કિંમત શોધો.

નિશ્ચિત સંકલનની વ્યાખ્યા દ્વારા, $\lim _{n \rightarrow \infty}\left(\frac{1^4}{1^5+n^5}+\frac{2^4}{2^5+n^5}+\frac{3^4}{3^5+n^5}+\ldots+\frac{n^4}{n^5+n^5}\right)$ નું મૂલ્ય શોધો.

સરવાળાની મર્યાદા તરીકે નીચેના નિશ્ચિત સંકલનનું મૂલ્ય શોધો:
$\int_{2}^{3} x^{2} d x$

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

$\lim _{n \rightarrow \infty}\left[\frac{n+3}{n^2+1^2}+\frac{n+6}{n^2+2^2}+\frac{n+9}{n^2+3^2}+\ldots+\frac{2}{n}\right]=$

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