 Solution

ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2013/discrete-mathematics-probability-and-continuous-mathematics/5/b/iii/solution

For

N

T_{N} = \frac{1}{N} [\frac{1}{2} + \sum_{n = 1}^{N - 1} (\frac{n}{N})^{2}] = \frac{1}{N ^{3}} \sum_{n = 1}^{N - 1} n^{2} + \frac{1}{2 N} .

Adding the missing

N^{2} / N^{3}

term therefore gives the exact relation

\frac{1}{N ^{3}} \sum_{n = 1}^{N} n^{2} = T_{N} + \frac{1}{2 N} .

Since

x^{2}

is continuous, its Riemann sums, and hence its endpoint-averaged trapezoidal sums, converge to its definite integral. Thus

T_{N} \to 1/3

and

1/ (2 N) \to 0

N \to \infty lim \frac{1}{N ^{3}} n = 1 \sum N n^{2} = \frac{1}{3} .

No pre-existing formula for the sum of squares is needed.

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