$\mathop {\lim }\limits_{n \to \infty } \,\sum\limits_{r = 0}^n {\frac{n}{{{{\left( {2r + n} \right)}^2}}}} $ का मान ज्ञात कीजिए।

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

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Similar Questions

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

$\mathop {\lim }\limits_{n \to \infty } {\left\{ {\left( {1 + \frac{{{1^2}}}{{{n^2}}}} \right)\left( {1 + \frac{{{2^2}}}{{{n^2}}}} \right)\left( {1 + \frac{{{3^2}}}{{{n^2}}}} \right) \dots \left( {1 + \frac{{{{(n - 1)}^2}}}{{{n^2}}}} \right)} \right\}^{1/n}}$ का मान ज्ञात कीजिए:

$\lim _{n \rightarrow \infty} n\left[\frac{1}{3 n^2+8 n+4}+\frac{1}{3 n^2+16 n+16}+\ldots+\frac{1}{15 n^2}\right]=$

$n$ के पर्याप्त बड़े मान के लिए,प्रथम $n$ धनात्मक पूर्णांकों के वर्गमूलों का योग,अर्थात $\sqrt{1} + \sqrt{2} + \sqrt{3} + \dots + \sqrt{n}$,लगभग किसके बराबर है?

$\lim _{n \rightarrow \infty} \frac{1}{n^{k+1}}\left[2^k+4^k+6^k+\ldots+(2 n)^k\right]=$

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