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