 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-oxford/2015/prelims/mathematics-ii/3/a/iii/solution

The finite geometric series identity, obtained by subtracting

t

times the sum from the sum itself, gives

S_{N} = \sum_{n = 0}^{N} t^{n} = \frac{1 - t ^{N + 1}}{1 - t} .

For

0 < t < 1

t^{N + 1}

tends to zero by Question 2(a)(iii). Therefore

n = 0 \sum \infty t^{n} = \frac{1}{1 - t} .

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