 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2019/paper-1/19/a/i/solution

For every integer

n

cos ((2 n - 1) π) = - 1

. The series is therefore the negative of the harmonic series:

\sum_{n = 1}^{N} \frac{c o s (( 2 n - 1 ) π )}{n} = - \sum_{n = 1}^{N} \frac{1}{n} .

For a direct divergence argument, each block

2^{j - 1} < n \leq 2^{j}

contributes at least

2^{j - 1} / 2^{j} = 1/2

to the positive sum. Its partial sums grow without bound, so

the given series diverges to - \infty.

Despite the odd multiples of

π

, the signs do not alternate.

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