$\mathop {\lim }\limits_{n \to \infty } \left[ {\frac{1}{n} + \frac{1}{{n + 1}} + \frac{1}{{n + 2}} + \dots + \frac{1}{{2n}}} \right] = $

  • A
    $0$
  • B
    $\log_e 4$
  • C
    $\log_e 3$
  • D
    $\log_e 2$

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Similar Questions

$\mathop {\lim }\limits_{n \to \infty } \left( {\frac{{{{\left( {n + 1} \right)}^{1/3}}}}{{{n^{4/3}}}} + \frac{{{{\left( {n + 2} \right)}^{1/3}}}}{{{n^{4/3}}}} + \dots + \frac{{{{\left( {2n} \right)}^{1/3}}}}{{{n^{4/3}}}}} \right)$ ની કિંમત શોધો.

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

$\lim _{n \rightarrow \infty} \left( \frac{1}{\sqrt{4n^2-1}} + \frac{1}{\sqrt{4n^2-4}} + \dots + \frac{1}{\sqrt{4n^2-n^2}} \right)$ ની કિંમત શોધો.

$\mathop {\lim }\limits_{n \to \infty } \frac{{1 + {2^4} + {3^4} + .... + {n^4}}}{{{n^5}}} - \mathop {\lim }\limits_{n \to \infty } \frac{{1 + {2^3} + {3^3} + .... + {n^3}}}{{{n^5}}} = $

$\lim _{n}$ ${\rightarrow \infty} \frac{1}{n}\left(\frac{1}{e^{1 / n}}+\frac{1}{e^{2 / n}}+\frac{1}{e^{3 / n}}+\ldots+\frac{1}{e^{2n/n}}\right)=$

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