 Solution

ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2019/discrete-mathematics-continuous-mathematics-and-probability-resit/3/a/iv/solution

True. The logarithm identity gives

ln (e^{n} + n) - n = ln (1 + n e^{- n}) .

For

n \geq 1

e^{n} \geq n

, so

0 \leq n e^{- n} \leq 1

. By monotonicity of the natural logarithm, the expression lies between

0

and

ln 2

. It is therefore

O (1)

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