Science:Math Exam Resources/Courses/MATH101/April 2017/Question 05 (b)/Solution 1

In fact, we can change expression in the question to following:

$$\sum _{i=1}^{n}{\frac {n}{n^{2}+i^{2}}}=\sum _{i=1}^{n}{\frac {n^{2}}{n^{2}+i^{2}}}\cdot {\frac {1}{n}}=\sum _{i=1}^{n}{\frac {1}{1+({\frac {i}{n}})^{2}}}\cdot {\frac {1}{n}}.$$

From this form, we can easily see every element in the Riemman Sum that

$$f(x)={\frac {1}{1+x^{2}}},\ \ x_{i}={\frac {i}{n}},\ \ \Delta x=|x_{i}-x_{i-1}|={\frac {1}{n}}.$$

Then this limit becomes

$$\lim _{n\rightarrow \infty }\sum _{i=1}^{n}{\frac {n}{n^{2}+i^{2}}}=\lim _{n\rightarrow \infty }\sum _{i=1}^{n}f(x_{i})\Delta x=\int _{x_{0}}^{x_{n}}f(x)dx=\int _{0}^{1}{\frac {1}{1+x^{2}}}dx=\arctan x{\bigg |}_{0}^{1}=\color {blue}{\frac {\pi }{4}}.$$

Answer: ${\frac {\pi }{4}}.$