 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-oxford/2017/prelims/mathematics-v-resit/question-4/c/solution

p = 3

and

x = 0

0 = \frac{1}{3} - \frac{1}{6} - 6 \sum_{n \geq 1 n \neq = 3} \frac{( - 1 ) ^{n}}{n ^{2} - 9} .

\sum_{n \geq 1 n \neq = 3} \frac{( - 1 ) ^{n}}{n ^{2} - 9} = \frac{1}{36} .

The

n = 1

and

n = 2

terms are

1/8

and

- 1/5

, respectively. Subtracting them gives the required absolutely convergent series:

n = 4 \sum \infty \frac{( - 1 ) ^{n}}{n ^{2} - 9} = \frac{1}{36} - \frac{1}{8} + \frac{1}{5} = \frac{37}{360} .

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