Past exam of the graph Theory and Probabilistic Combinatorics course of the University of Oxford / 2008 / B / 5 / a

Table of contents
- i a
  - Solution i
- ii a
  - Solution ii

i

Solution

For

k \geq 1

, monotonicity of the logarithm gives

lo g (k!) = \sum_{j = 1}^{k} lo g j \geq \int_{1}^{k} lo g t d t = k lo g k - k + 1 \geq k lo g k - k .

Thus

k! \geq (k / e)^{k}

. Bounding the numerator of the binomial coefficient by

n^{k}

yields

(k n) \leq \frac{n ^{k}}{k !} \leq (e n / k)^{k} .

ii

Solution

x - lo g (1 + x)

x / (1 + x)

, is zero at zero, and is nonnegative for every

x > - 1

. Therefore

lo g (1 + x) \leq x

. Multiplying by

m > 0

and exponentiating gives

(1 + x)^{m} \leq e^{x m} .

Equality occurs at

x = 0

.

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