<s>[INST] How to find $\lim\limits_{n\to\infty}\sum\limits_{j=1}^{n^2}\frac{n}{n^2+j^2}$ find the limit value

$$\lim_{n\to\infty}\sum_{j=1}^{n^2}\dfrac{n}{n^2+j^2}$$

this following is my methods:

let $$S_{n}=\sum_{j=1}^{n^2}\dfrac{n}{n^2+j^2}=\sum_{j=1}^{n^2}\dfrac{1}{1+\left(\dfrac{j}{n}\right)^2}\dfrac{1}{n}$$
  since
  $$\int_{\dfrac{j}{n}}^{\dfrac{j+1}{n}}\dfrac{dx}{1+x^2}<\dfrac{1}{1+\left(\dfrac{j}{n}\right)^2}\cdot\dfrac{1}{n}<\int_{\dfrac{j-1}{n}}^{\dfrac{j}{n}}\dfrac{dx}{1+x^2}$$
  so
  $$\int_{\dfrac{1}{n}}^{\dfrac{n^2+1}{n}}\dfrac{dx}{1+x^2}<S_{n}<\int_{0}^{n}\dfrac{dx}{1+x^2}$$

and note

$$\lim_{n\to\infty}\int_{\dfrac{1}{n}}^{\dfrac{n^2+1}{n}}\dfrac{dx}{1+x^2}=\lim_{n\to\infty}\int_{0}^{n}\dfrac{dx}{1+x^2}=\int_{0}^{infty}\dfrac{dx}{1+x^2}=\dfrac{\pi}{2}$$
  so
  $$\lim_{n\to\infty}\sum_{j=1}^{n^2}\dfrac{n}{n^2+j^2}=\dfrac{\pi}{2}$$

I think this problem have other nice methods? Thank you 
and follow other methods 

$$\lim_{n\to\infty}\sum_{j=1}^{n^2}\dfrac{n}{n^2+j^2}=\lim_{n\to\infty}\int_{0}^{n}\dfrac{1}{1+x^2}dx=\dfrac{\pi}{2}$$
  But there is a book say This methods is wrong,why, and where is wrong? Thank you 

 [/INST] Here is a method in the same spirit as what you have done. The main difference is that we explicitly estimate the difference between your sum and $\int_0^n\frac{{\rm d}x}{1+x^2}$.
Since $f$ is monotonely decreasing we have (see for example integral test on Wikipedia)
$$\left|\sum_{j=1}^n \frac{n}{n^2+(j+nm)^2} - \int_{m}^{m+1}\frac{{\rm d}x}{1+x^2}\right| < \frac{1}{n(1 + (m+1)^2)}$$
Using this we can estimate the difference between your sum and the integral of $\frac{1}{1+x^2}$ over $[0,n]$, which can be written $\sum_{m=0}^{n-1}\int_0^1\frac{{\rm d}x}{1+(x+m)^2}$, as
$$\left|\sum_{j=1}^{n^2}\frac{n}{n^2 + j^2} - \int_0^n\frac{{\rm d}x}{1+x^2}\right|  \leq \sum_{m=0}^{n-1} \left|\frac{1}{n}\sum_{j=1}^n \frac{1}{1+\left(\frac{j}{n}+m\right)^2} - \int_{0}^{1}\frac{{\rm d}x}{1+(x+m)^2}\right| \\\leq \frac{1}{n}\sum_{m=1}^{n} \frac{1}{1+m^2} \leq \frac{1}{n}\sum_{m=1}^\infty \frac{1}{1+m^2}$$
This last sum can be evaluated as $\frac{\pi\coth(\pi)-1}{2}$, however all we need for this argument is that it's finite. This can be proved for example by comparison to $\sum_{m=1}^\infty \frac{1}{m^2} = \frac{\pi^2}{6}$. Taking $n\to\infty$ the result follows
$$\lim_{n\to\infty}\sum_{j=1}^{n^2}\frac{n}{n^2 + j^2} = \lim_{n\to\infty}\int_0^n\frac{{\rm d}x}{1+x^2} = \frac{\pi}{2}$$
</s>