 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-resit/4/a/iii/solution

Let

S_{N} = N^{- 3} \sum_{n = 1}^{N} n^{2}

. The expression in part (ii) gives

T_{N} = 3 (S_{N} - \frac{1}{N}) + \frac{3}{2 N} = 3 S_{N} - \frac{3}{2 N} .

Because

3 x^{2}

[0, 1]

1

∣ T_{N} - 1∣ \leq 1/ (2 N^{2})

, since

f^{''} = 6

. Therefore

N \to \infty lim S_{N} = N \to \infty lim (\frac{T _{N}}{3} + \frac{1}{2 N}) = \frac{1}{3} .

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