Infinite!

Calculus Level 4

7 9 × 26 28 × 63 65 × × k 3 1 k 3 + 1 × = ? \large \dfrac79 \times \dfrac{26}{28} \times \dfrac{63}{65} \times \cdots \times \dfrac{k^3-1}{k^3+1} \times \cdots = \, ?

Give your answer to 2 decimal places.


The answer is 0.67.

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1 solution

P = k = 2 k 3 1 k 3 + 1 = k = 2 ( k 1 ) ( k 2 + k + 1 ) ( k + 1 ) ( k 2 k + 1 ) = k = 2 ( k 1 ) k = 2 ( k 2 + k + 1 ) k = 2 ( k + 1 ) k = 2 ( k 2 k + 1 ) = k = 1 k × k = 2 ( k 2 + k + 1 ) k = 3 k × k = 1 ( k 2 + k + 1 ) We note that ( k + 1 ) 2 ( k + 1 ) + 1 = k 2 + k + 1 = 1 × 2 1 2 + 1 + 1 = 2 3 0.67 \begin{aligned} P & = \prod_{k=2}^\infty \frac{k^3-1}{k^3+1} \\ & = \prod_{k=2}^\infty \frac{(k-1)(k^2+k+1)}{(k+1)(k^2-k+1)} \\ & = \frac{\displaystyle \prod_{k=2}^\infty (k-1) \prod_{k=2}^\infty (k^2+k+1)}{\displaystyle \prod_{k=2}^\infty (k+1) \prod_{k=2}^\infty (k^2-k+1)} \\ & = \frac{\displaystyle \prod_{k=1}^\infty k \times \prod_{k=2}^\infty (k^2+k+1)}{\displaystyle \prod_{k=3}^\infty k \times \prod_{k=\color{#3D99F6}{1}}^\infty (\color{#3D99F6}{k^2+k+1})} \quad \quad \small \color{#3D99F6}{\text{We note that }(k+1)^2-(k+1)+1 = k^2 + k +1} \\ & = \frac{1\times 2}{1^2+1+1} = \frac{2}{3} \approx \boxed{0.67} \end{aligned}

Very Gooooooood solution!!!!!!!!!

Ayush G Rai - 5 years, 1 month ago

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