$\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) = $

  • A
    $\frac{\pi }{4} + \ln 2$
  • B
    $\frac{\pi }{4} - \ln 2$
  • C
    $\frac{\pi }{4} + \frac{1}{2}\ln 2$
  • D
    $\frac{\pi }{4} - \frac{1}{2}\ln 2$

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Similar Questions

$\mathop {Lim}\limits_{n \to \infty } \frac{\pi }{{6n}}\left[ {{{\sec }^2}\left( {\frac{\pi }{{6n}}} \right) + {{\sec }^2}\left( {2 \cdot \frac{\pi }{{6n}}} \right) + \dots + {{\sec }^2}\left( {(n - 1)\frac{\pi }{{6n}}} \right) + \frac{4}{3}} \right]$ ની કિંમત કેટલી થાય?

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

$\lim _{n \rightarrow \infty} \frac{1}{n^3} \sum_{k=1}^n (k^2 x)$ ની કિંમત શોધો.

આપેલ છે કે $\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^{n p} f\left(\frac{r}{n}\right)=\int_0^p f(x) d x$. જો $f: R \rightarrow R$ એ $f(x)=x^2+2$ દ્વારા વ્યાખ્યાયિત હોય,તો $\lim _{n \rightarrow \infty} \frac{3}{n}\left[f\left(\frac{7}{n}\right)+f\left(\frac{14}{n}\right)+f\left(\frac{21}{n}\right)+\ldots+f(7)\right]=$

$\lim _{n \rightarrow \infty}\left\{\frac{1}{\sqrt{4 n^2-1^2}}+\frac{1}{\sqrt{4 n^2-2^2}}+\frac{1}{\sqrt{4 n^2-3^2}}+\dots+\frac{1}{\sqrt{4 n^2-n^2}}\right\}=$

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