 Solution

ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2015/discrete-mathematics-continuous-mathematics-and-probability-resit/2/b/ii/solution

n = 2

1 + 1/ 2 > 2

, since multiplying by

2

reduces this to

2 + 1 > 2

. Suppose the claimed inequality holds for some

n \geq 2

. The identity

n + 1 - n = \frac{1}{n + 1 + n} < \frac{1}{n + 1}

then gives

\sum_{k = 1}^{n + 1} \frac{1}{k} > n + \frac{1}{n + 1} > n + 1 .

n < k = 1 \sum n \frac{1}{k} (n \geq 2) .

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