Question

Solve $\underset{n\to \infty }{lim}\frac{1}{n}\sum _{r=0}^{2n-1}\frac{n^2}{n^2+4r^2}$

Answer 1

Finding the value of $\underset{n\to \infty }{lim}\frac{1}{n}\sum _{r=0}^{2n-1}\frac{n^2}{n^2+4r^2}$

Given: $\underset{n\to \infty }{lim}\frac{1}{n}\sum _{r=0}^{2n-1}\frac{n^2}{n^2+4r^2}$

$$\begin{gathered} L=\underset{n\to \infty }{\lim }\frac{1}{n}\sum _{r=0}^{2n-1}\frac{n^2}{n^2+4r^2} \\ \underset{}{L=\underset{n\to \infty }{\lim }}\frac{1}{n}\sum _{r=0}^{2n-1}\frac{1}{1+\frac{4r^2}{n^2}} \end{gathered}$$

Replace

$$\begin{gathered} \frac{r}{n}\to x,\frac{1}{n}\to \frac{dy}{dx} \\ \text{lower limit =0} \\ \mathrm{U}\text{pper limit}=\underset{\mathrm{n}\to \infty }{\lim }\frac{2\mathrm{n}-1}{\mathrm{n}}=2 \end{gathered}$$

Now, on integration we get

$$\begin{gathered} \begin{array}{l}L={\int }_0^2\frac{1}{1+(2x{)}^2}dx\end{array} \\ L=\overline{)\frac{{\tan }^{-1}\left(2x\right)}{2}}\begin{array}{c}2\\ \\ 0\end{array} \\ L=\left(\frac{1}{2}\right){\tan }^{-1}4–0 \end{gathered}$$

Hence, the value of $\underset{\mathbf{n}\mathbf{\to }\mathbf{\infty }}{\mathbf{lim}}\frac{\mathbf{1}}{\mathbf{n}}\mathbf{\sum }_{\mathbf{r}\mathbf{=}\mathbf{0}}^{\mathbf{2}\mathbf{n}\mathbf{-}\mathbf{1}}\frac{{\mathbf{n}}^{\mathbf{2}}}{{\mathbf{n}}^{\mathbf{2}}\mathbf{+}\mathbf{4}{\mathbf{r}}^{\mathbf{2}}}$is $\mathbf{\left(}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{\right)}{\mathbf{tan}}^{\mathbf{-}\mathbf{1}}\mathbf{4}$.