 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2025/paper-1/19/d/solution

n

, including

n = 1

h (t) = ln (1 + t) - \frac{2 t}{2 + t}, t \geq 0.

Then

h (0) = 0

, while direct differentiation gives

h^{'} (t) = \frac{1}{1 + t} - \frac{4}{( 2 + t ) ^{2}} = \frac{t ^{2}}{( 1 + t ) ( 2 + t ) ^{2}} > 0 (t > 0) .

Thus

h (t) > 0

for

t > 0

. Setting

t = 1/ n

in the expression from part (b) gives

u_{n} = \frac{2 + t}{2 t} [ln (1 + t) - \frac{2 t}{2 + t}] = \frac{2 + t}{2 t} h (t) > 0.

Therefore

u_{n} > 0 for every n \geq 1

, and

(a_{n})

is a strictly increasing sequence.

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