$\underset{n\to \infty }{\mathop{\lim }}\,\frac{{{1}^{99}}+{{2}^{99}}+{{3}^{99}}+......{{n}^{99}}}{{{n}^{100}}}=$

  • A
    $\frac{9}{100}$
  • B
    $\frac{1}{100}$
  • C
    $\frac{1}{99}$
  • D
    $\frac{1}{101}$

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Similar Questions

$\mathop {\lim }\limits_{n \to \infty } \left( {\frac{{n + 1}}{{{n^2} + {1^2}}} + \frac{{n + 2}}{{{n^2} + {2^2}}} + \frac{{n + 3}}{{{n^2} + {3^2}}} + \dots + \frac{1}{n}} \right) = $

$\lim _{n \rightarrow \infty} \sum_{k=1}^n \frac{n^3}{(n^2+k^2)(n^2+3k^2)}$ નું મૂલ્ય શું છે?

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

લક્ષની કિંમત શોધો: $\lim _{n \rightarrow \infty} \frac{3}{n}\left\{1+\sqrt{\frac{n}{n+3}}+\sqrt{\frac{n}{n+6}}+\sqrt{\frac{n}{n+9}}+\ldots+\sqrt{\frac{n}{n+3(n-1)}}\right\}$

$\mathop {Lim}\limits_{n \to \infty } \,\,\sum\limits_{k = 1}^n {\frac{n}{{{n^2} + {k^2}{x^2}}}} $,$x > 0$ ની કિંમત શોધો.

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