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

  • A
    $\pi / 4$
  • B
    $\tan ^{-1} 3$
  • C
    $\tan ^{-1} 2$
  • D
    $\pi / 2$

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Similar Questions

निश्चित समाकल की परिभाषा के अनुसार,$\lim _{n \rightarrow \infty}\left(\frac{1}{\sqrt{n^2-1^2}}+\frac{1}{\sqrt{n^2-2^2}}+\ldots+\frac{1}{\sqrt{n^2-(n-1)^2}}\right)$ का मान किसके बराबर है?

$\lim _{n \rightarrow \infty} \frac{1}{n} \left\{ \sec ^{2} \frac{\pi}{4 n} + \sec ^{2} \frac{2 \pi}{4 n} + \ldots + \sec ^{2} \frac{n \pi}{4 n} \right\}$ का मान ज्ञात कीजिए।

$\lim _{n \rightarrow \infty}\left[\frac{n+3}{n^2+1^2}+\frac{n+6}{n^2+2^2}+\frac{n+9}{n^2+3^2}+\ldots+\frac{2}{n}\right]=$

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

निम्नलिखित कथनों में से:
$(S1): \lim _{n \rightarrow \infty} \frac{1}{n^2}(2+4+6+\ldots+2n)=1$
$(S2): \lim _{n \rightarrow \infty} \frac{1}{n^{16}}(1^{15}+2^{15}+3^{15}+\ldots+n^{15})=\frac{1}{16}$

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