Q115 / 160 · TS EAMCET 2019 (03 May Shift 1) · Mathematics

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

  • Correct answer
    A
    $\frac{1}{2} \log 2$
  • B
    $2 \log 2$
  • C
    $\frac{1}{3} \log 2$
  • D
    $3 \log 2$

Solution

We have, $$ \begin{aligned} & I=\lim _{n \rightarrow \infty} \sum_{k=1}^n \frac{k}{n^2+k^2}=\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{k=1}^n \frac{k / n}{1+\left(\frac{k}{n}\right)^2} \\ \Rightarrow & I=\int_0^1 \frac{x}{1+x^2} d x=\frac{1}{2}\left[\log \left(1+x^2\right)\right]_0^1=\frac{1}{2} \log 2 \end{aligned} $$

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

Relevant Questions