 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/b/solution

Cancellation of the factorials in the consecutive ratio gives

\frac{a _{n + 1}}{a _{n}} = e^{- 1} (1 + \frac{1}{n})^{n + 1/2}, u_{n} = (n + \frac{1}{2}) ln (1 + \frac{1}{n}) - 1.

t = 1/ n

1

gives

ln (1 + t) / t \to 1

. Therefore

u_{n} = (1 + \frac{t}{2}) \frac{l n ( 1 + t )}{t} - 1 ⟶ 0 .

Thus the necessary condition from the term test for divergence is satisfied. On its own it does not prove that

\sum u_{n}

converges.

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