2022TrialShort answer

Q16 (b) (ii)

3 marks

Question

Given
$S_n=\frac1{3^3}+\frac1{4^3}+\frac1{5^3}+\cdots+\frac1{n^3},\qquad n\ge3,$
use part (i), or otherwise, to prove that
$S_n<\frac1{12}.$

Try the question before revealing the answer.

Solution

By adding and cancelling,

S_n\leq\frac14-\frac13+\frac16-\frac12n+\frac1n-\frac1{2n+2}

S_n\leq\frac1{12}-\frac1{2n}+\frac1{2n+2}

S_n\leq\frac1{12}+\frac{-n-1+n}{2n(n+1)}

S_n\leq\frac1{12}-\frac1{2n(n+1)}

S_n\leq\frac1{12},\qquad\text{since }\frac{-1}{2n(n+1)}<0.

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Proof

The Nature of ProofUse formal proof language, contradiction, counter-examples and rigorous arguments involving numbers and inequalities.

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