Q103 / 180 · KCET 2024 · Mathematics

$\lim _{n \rightarrow \infty}\left(\frac{n}{n^2+1^2}+\frac{n}{n^2+2^2}+\frac{n}{n^2+3^2}+\ldots+\frac{1}{5 n}\right)=$

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

Solution

$\lim _{n \rightarrow \infty}\left(\frac{n}{n^2+1^2}+\frac{n}{n^2+2^2}+\frac{n}{n^2+3^2}+\ldots+\frac{1}{5 n}\right)$ $=\lim _{n \rightarrow \infty}\left(\frac{n}{n^2+1^2}+\frac{n}{n^2+2^2}+\frac{n}{n^2+3^2}+\ldots+\frac{n}{n^2+(2 n)^2}\right)$ $=\lim _{n \rightarrow \infty} \sum_{r=1}^{2 n} \frac{n}{n^2+r^2}$ $=\lim _{n \rightarrow \infty} \sum_{r=1}^{2 n} \frac{1}{n}\left(\frac{1}{1+\left(\frac{r}{n}\right)^2}\right)$ $=\int_0^2 \frac{d x}{1+x^2}=\tan ^{-1}(2)$

Hence, the correct answer is option: 'C'.

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