 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/c/solution

For

n > 1

, the Taylor series of the natural logarithm converges absolutely at

1/ n

ln (1 + \frac{1}{n}) = \sum_{j = 1}^{\infty} \frac{( - 1 ) ^{j + 1}}{j n ^{j}} .

Multiply this power series by

n + 1/2

and subtract

1

. The constant term cancels the subtracted

1

, and the coefficient of

1/ n

- 1/2 + 1/2 = 0

. For

k \geq 2

n^{- k}

\frac{( - 1 ) ^{k}}{k + 1} + \frac{( - 1 ) ^{k + 1}}{2 k} = (- 1)^{k} (\frac{1}{k + 1} - \frac{1}{2 k}) .

Hence

u_{n} = k = 2 \sum \infty \frac{( - 1 ) ^{k}}{n ^{k}} (\frac{1}{k + 1} - \frac{1}{2 k}) .

In particular

u_{n} = 1/ (12 n^{2}) - 1/ (12 n^{3}) + O (n^{- 4})

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