$\mathop {Limit}\limits_{n \to \infty } \frac{1}{n} \left[ 1 + \sqrt {\frac{n}{n + 1}} + \sqrt {\frac{n}{n + 2}} + \sqrt {\frac{n}{n + 3}} + \dots + \sqrt {\frac{n}{n + 3(n - 1)}} \right]$ has the value equal to

  • A
    $2(\sqrt{4} - 1)$
  • B
    $2(\sqrt{3} - 1)$
  • C
    $2$
  • D
    $4$

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Similar Questions

$\mathop {\lim }\limits_{n \to \infty } \frac{1}{{{n^2}}}\left[ {1\cos \frac{1}{{{n^2}}} + 2\cos \frac{4}{{{n^2}}} + 3\cos \frac{9}{{{n^2}}} + .... + 2n\cos 4} \right]$ is equal to-

$\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^{2 n} \frac{r}{\sqrt{n^2+r^2}}=$

$\mathop {\lim }\limits_{n \to \infty } \,\sum\limits_{r = 0}^n {\frac{n}{{{{\left( {2r + n} \right)}^2}}}} $ is equal to

$\lim _{n \rightarrow \infty}\left[\frac{1}{n^2} \sec ^2 \frac{1}{n^2}+\frac{2}{n^2} \sec ^2 \frac{4}{n^2}+\frac{3}{n^2} \sec ^2 \frac{9}{n^2}+\ldots+\frac{n}{n^2} \sec ^2 \frac{n^2}{n^2}\right]=$

The value of $\mathop {\lim }\limits_{n \to \infty } \left[ {\frac{n}{{1 + {n^2}}} + \frac{n}{{4 + {n^2}}} + \frac{n}{{9 + {n^2}}} + .... + \frac{1}{{2n}}} \right]$ is equal to

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