$$\lim_{n \to \infty} \left(\frac{n}{n^2 + 1^2} + \frac{n}{n^2 + 2^2} + \frac{n}{n^2 + 3^2} + \ldots + \frac{1}{5n^2}\right)$$ is equal to

We have to evaluate the limit

$$L=\lim_{n\to\infty}\Bigg(\frac{n}{\,n^{2}+1^{2}\,}+\frac{n}{\,n^{2}+2^{2}\,}+\frac{n}{\,n^{2}+3^{2}\,}+\,\ldots\,+\frac{n}{\,n^{2}+(2n)^{2}\,}\Bigg).$$

(The last term has the index $$k=2n$$ and therefore its numerator is still $$n$$. This keeps the pattern uniform and allows us to treat the whole bracket as one summation.)

First we rewrite every term in a way that makes a Riemann-sum structure obvious. For a generic index $$k$$ (with $$1\le k\le 2n$$) the $$k^{\text{th}}$$ term is

$$\frac{n}{\,n^{2}+k^{2}\,}.$$

Dividing numerator and denominator by $$n^{2}$$ we get

$$\frac{n}{\,n^{2}+k^{2}\,}= \frac{n}{n^{2}\bigl(1+(k/n)^{2}\bigr)} =\frac{1}{n}\;\frac{1}{\,1+(k/n)^{2}\,}.$$

So the entire bracket can be rewritten as

$$\sum_{k=1}^{2n}\frac{n}{\,n^{2}+k^{2}\,} =\sum_{k=1}^{2n}\frac{1}{n}\;\frac{1}{\,1+(k/n)^{2}\,}.$$

Now set

$$x_k=\frac{k}{n}\quad\text{and}\quad\Delta x=\frac{1}{n}.$$

Because $$k$$ runs from $$1$$ to $$2n$$, the variable $$x_k$$ runs from

$$x_1=\frac{1}{n}\longrightarrow x_{2n}=\frac{2n}{n}=2.$$

Hence the sum can be recognised as

$$\sum_{k=1}^{2n} \bigl[f(x_k)\bigr]\,\Delta x, \quad\text{where}\quad f(x)=\frac{1}{1+x^{2}}.$$

By the definition of the definite integral (Riemann sum), we have the formula

$$\int_{a}^{b}f(x)\,dx=\lim_{n\to\infty}\sum_{k=1}^{n}f(x_k^{\!*})\,\Delta x,$$ where each $$x_k^{\!*}$$ lies in the $$k^{\text{th}}$$ sub-interval of width $$\Delta x$$. In our sum $$x_k=k/n$$ is a perfectly acceptable choice of such sample points.

Therefore, taking $$a=0$$, $$b=2$$ and $$f(x)=\dfrac{1}{1+x^{2}}$$, the bracket tends to

$$\int_{0}^{2}\frac{1}{1+x^{2}}\,dx.$$

We recall the standard antiderivative

$$\int\frac{1}{1+x^{2}}\,dx=\tan^{-1}(x)+C.$$

Applying the limits $$0$$ and $$2$$ we get

$$\int_{0}^{2}\frac{1}{1+x^{2}}\,dx =\Bigl[\tan^{-1}(x)\Bigr]_{0}^{2} =\tan^{-1}(2)-\tan^{-1}(0).$$

Since $$\tan^{-1}(0)=0$$, this reduces to

$$\tan^{-1}(2).$$

So the limit we were asked to find is exactly the real number $$\tan^{-1}(2)$$.

Hence, the correct answer is Option B.