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Question Number 9004 by mrW last updated on 12/Nov/16

prove

\frac{1}{2} \cdot \frac{3}{4} \cdot \frac{5}{6} \cdot \cdot \cdot \cdot \cdot \frac{2 n - 1}{2 n} ⩽ \frac{1}{\sqrt{3 n + 1}}

Answered by mrW last updated on 15/Nov/16

Using mathematical induction

prove P (n) = \underset{k = 1}{\prod^{n}} \frac{2 k - 1}{2 k} ⩽ \frac{1}{\sqrt{3 n + 1}}

for n = 1

P (1) = \frac{1}{2} \overset{!}{⩽} \frac{1}{\sqrt{3 \cdot 1 + 1}} = \frac{1}{\sqrt{4}} = \frac{1}{2}

{it}^{'} s true .

suppose {it}^{'} s true for n, i . e .

P (n) = \underset{k = 1}{\prod^{n}} \frac{2 k - 1}{2 k} ⩽ \frac{1}{\sqrt{3 n + 1}}

P (n + 1) = \frac{2 (n + 1) - 1}{2 (n + 1)} P (n)

= \frac{2 n + 1}{2 n + 2} P (n) ⩽ \frac{2 n + 1}{2 n + 2} \cdot \frac{1}{\sqrt{3 n + 1}}

= \frac{2 n + 1}{2 n + 2} \cdot \frac{\sqrt{3 n + 4}}{\sqrt{3 n + 1}} \cdot \frac{1}{\sqrt{3 (n + 1) + 1}}

= \frac{\sqrt{{(2 n + 1)}^{2} \cdot (3 n + 4)}}{\sqrt{{(2 n + 2)}^{2} \cdot (3 n + 1)}} \cdot \frac{1}{\sqrt{3 (n + 1) + 1}}

= \frac{\sqrt{({4 n}^{2} + 4 n + 1) \cdot (3 n + 4)}}{\sqrt{({4 n}^{2} + 8 n + 4) \cdot (3 n + 1)}} \cdot \frac{1}{\sqrt{3 (n + 1) + 1}}

= \frac{\sqrt{{12 n}^{3} + {28 n}^{2} + 19 n + 4}}{\sqrt{{12 n}^{3} + {28 n}^{2} + 20 n + 4}} \cdot \frac{1}{\sqrt{3 (n + 1) + 1}}

⩽ \frac{1}{\sqrt{3 (n + 1) + 1}}

so {it}^{'} s true for all n ⩾ 1

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