To find the sum of this series I need to find the sum of this series $$
\frac{n}{n^2}+\frac{n}{n^2+1^2}+\frac{n}{n^2+2^2}+\ldots+\frac{n}{n^2+(n-1)^2}
$$
Here general term $=\frac{n}{n^2+r^2}, r=0,1,2, \ldots$
$$
\begin{aligned}
& =\frac{1}{n}\left[\frac{1}{1+\left(\frac{r}{n}\right)^2}\right]  \\
& =\frac{d x}{1+x^2}\left[\text { Replace } \frac{1}{n} \text { by } d x \text { and } \frac{r}{n} \text { by } x\right]
\end{aligned}
$$
Required sum $=\int_0^1 \frac{d x}{1+x^2}=\left.\tan ^{-1} x\right|_0 ^1=\frac{\pi}{4}$
In this how to get the lower limit and upper limit values $0$ and $1$ respectively ???
Thanks in advance.

Upper limit= $\lim_{n\to\infty} \frac{n}{n}$,
Lower limit= $\lim_{n\to\infty} \frac{0}{n}$.
That gives $1$ and $0$.
In general if this type of sum comes up with the lower limit being $a_n$ and upper being $ b_n$. Then Upper limit of integral = $\lim_{n\to\infty} \frac{b_n}{n}$,
Lower limit of integral = $\lim_{n\to\infty} \frac{a_n}{n}$.
