 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-oxford/2023/prelims/mathematics-ii/2/c/iii/solution

Let

u (x) = (lo g x) x^{- 2/3}

u^{'} (x) = x^{- 5/3} (1 - \frac{2}{3} lo g x),

which is negative for

x > e^{3/2}

. Also

u (x) \to 0

, since a positive power eventually dominates the natural logarithm. Thus the alternating series test proves convergence of

\sum (- 1)^{n} u (n)

For

n \geq 3

u (n) \geq n^{- 2/3}

, whose p-series diverges. The comparison test therefore rules out absolute convergence. The answer is conditional convergence.

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