 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2023/paper-1/19/b/ii/solution

For

u_{k} = a^{2 k + 1} / (2 k + 1) > 0

, the consecutive ratio is

\frac{u _{k + 1}}{u _{k}} = a^{2} \frac{2 k + 1}{2 k + 3} ⟶ a^{2} .

Thus the ratio test proves convergence when

0 < a < 1

and divergence when

a > 1

. At

a = 1

, the series is

\sum_{k \geq 0} 1/ (2 k + 1)

, which diverges by comparison with

\frac{1}{2} \sum_{k \geq 0} 1/ (k + 1)

. Therefore

convergent for 0 < a < 1; divergent for a \geq 1.

All terms are nonnegative, so convergence here is also absolute convergence.

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