Answer
$\lim _{n \rightarrow+\infty} \sum_{k=1}^{n} \frac{n}{k^{2}+n^{2}}=\int_{0}^{1} \frac{1}{x^{2}+1} d x=\frac{\pi}{4}$
Work Step by Step
Using the rules of Riemann sums:
\[
\begin{aligned}
\lim _{n \rightarrow+\infty} \sum_{k=1}^{n} \frac{n}{k^{2}+n^{2}} &=\lim _{n \rightarrow+\infty} \sum_{k=1}^{n} \frac{n}{\left(1+\left(\frac{k}{n}\right)^{2}\right)n^{2}} \\
&=\lim _{n \rightarrow+\infty} \sum_{k=1}^{n} \frac{\frac{1}{n}}{\left(\frac{k}{n}\right)^{2}+1} \\
&=\lim _{n \rightarrow+\infty} \sum_{k=1}^{n} \frac{1}{\left(\frac{k}{n}\right)^{2}+1} \cdot \frac{1}{n} \\
&=\lim _{n \rightarrow+\infty} \sum_{k=1}^{n} \Delta x_{k} f\left(x_{k}^{*}\right)
\end{aligned}
\]
$x_{k}^{*}=\frac{k}{n}, f(x)=\frac{1}{x^{2}+1}$ and $\Delta x_{k}=\frac{1}{n}=\frac{-0+1}{n}$
$\therefore \lim _{n \rightarrow+\infty} \sum_{k=1}^{n} \frac{n}{k^{2}+n^{2}}=\int_{0}^{1} \frac{1}{x^{2}+1} d x$
$\int_{0}^{1} \frac{1}{x^{2}+1} d x=\left.\tan ^{-1} x\right|_{0} ^{1}=-\tan ^{-1} 0+\tan ^{-1} 1=\frac{\pi}{4}$
