 Solution

ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2015/discrete-mathematics-continuous-mathematics-and-probability/6/a/ii/solution

Write the sum in the question as the right Riemann sum

S_{N} = \sum_{j = 1}^{N} \frac{N}{N ^{2} + j ^{2}} = \frac{1}{N} \sum_{j = 1}^{N} \frac{1}{1 + ( j / N ) ^{2}} .

Relative to the composite trapezoidal rule, the only change is in the two endpoint weights:

S_{N} - T_{N} = \frac{1}{2 N} (f (1) - f (0)) = - \frac{1}{4 N} .

f (x) = 1/ (1 + x^{2})

[0, 1]

, so

T_{N} \to I

. Since an antiderivative is

arctan x

I = arctan 1 - arctan 0 = \frac{π}{4} .

Thus

N \to \infty lim S_{N} = \frac{π}{4} .

 New to topics? Read the docs here!