How to compute the limit of $\frac{n}{n^2+1^2}+\frac{n}{n^2+2^2}+\frac{n}{n^2+3^2}+\cdots+\frac{n}{n^2+n^2}$ without using Riemann sums? 
How to compute the limit of $\frac{n}{n^2+1^2}+\frac{n}{n^2+2^2}+\frac{n}{n^2+3^2}+\cdots+\frac{n}{n^2+n^2}$ without using Riemann sums?

My Try: I have Solved It using Limit as a Sum (Reinman Sum of Integral.)
But I did not understand How can I solve it Using Sequeeze Theorem or any other way.

One may use the digamma function, from the standard $
\psi(z+1)-\psi(z)=\dfrac1z$ one obtains easily
$$
\psi(z+n+1)-\psi(z+1)=\sum_{k=1}^n\frac1{z+k},
$$ 
inserting $z:=in$ and considering imaginary parts gives
$$
\sum_{k=1}^n\frac{n}{n^2+k^2}=-\text{Im}\left[\psi(in+n+1)-\psi(in+1)\right]
$$ then one may recall that (see $6.3.18$ here)
$$
\psi(z)=\log z+O\left(\frac1z \right), \quad z \to \infty, \quad |\mathrm{arg}z|<\pi,
$$ which yields, as $n \to \infty$,

$$
\sum_{k=1}^n\frac{n}{n^2+k^2}=-\text{Im}\left[\log(1+i)-\log i \right]+O\left(\frac1n \right)=\frac\pi4+O\left(\frac1n \right) \to \frac\pi4.
$$


Remark. By considering more asymptotic terms in $\psi$, one gets,

$$
\sum_{k=1}^n\frac{n}{n^2+k^2}=\frac\pi4-\frac1{4n}-\frac1{24n^2}+\frac1{2016n^6}+O\left(\frac1{n^8} \right)
$$


$\displaystyle  \log z$ denotes the principal value of the logarithm defined by 
$$ \begin{align} 
  \displaystyle \log z  = \ln |z| + i \: \mathrm{arg}z, \quad -\pi <\mathrm{arg} z \leq \pi,\quad z \neq 0.
\end{align} 
$$
