There's only a minimal difference

Calculus Level 4

$\Large \prod_{r=3}^\infty \frac{r^3-8}{r^3+8} = \ ?$

Give your answer to 3 decimal places.

The answer is 0.2857.

1 solution

Chew-Seong Cheong
Apr 20, 2015

$\displaystyle \lim_{n\to \infty} {\prod_{r=3}^n\ {\frac{r^3-8}{r^3+8}}} = \prod_{r=3}^\infty\ {\frac{(r-2)(r^2+2r+4)}{(r+2)(r^2-2r+4)}} \\ = \dfrac{1\times 19}{5\times 7} \times \dfrac{2\times 28}{6\times12} \times \dfrac {3\times 39}{7\times 19} \times \dfrac {4\times 52}{8\times 28} \times \dfrac {5\times 67}{9\times 39} \times \dfrac {6\times 84}{10\times 52} \times ... \\ = \dfrac{1\times \color{#3D99F6}{\not{19}}}{\color{#D61F06}{\not{5}}\times 7} \times \dfrac{2\times \color{#3D99F6}{\not{28}}}{\color{#D61F06}{\not{6}}\times12} \times \dfrac {3\times \color{#3D99F6}{\not{39}}}{\color{#D61F06}{\not{7}}\times \color{#3D99F6}{\not{19}}} \times \dfrac {4\times \color{#3D99F6}{\not{52}}}{\color{#D61F06}{\not{8}}\times \color{#3D99F6}{\not{28}}} \times \dfrac {\color{#D61F06}{\not{5}}\times \color{#3D99F6}{\not{67}}}{\color{#D61F06}{\not{9}}\times \color{#3D99F6}{\not{39}}} \times \dfrac {\color{#D61F06}{\not{6}}\times \color{#3D99F6}{\not{84}}}{\color{#D61F06}{\not{10}}\times \color{#3D99F6}{\not{52}}} \times ... \\ = \dfrac {1}{7} \times \dfrac {2}{12} \times 3 \times 4 = \boxed{\dfrac{2}{7}}$

Moderator note:

This is wrong. Where did 67 and 84 cancel out with? The problem with your solution is that no matter how matter terms you write, you will always a have 2 terms that can't be cancelled out with.

Hint: Telescoping Product. Is there an equivalent form of $r^2 + 2r + 4$ and $r^2-2r + 4$ ?

@Chew-Seong Cheong What should we do if some non perfect cube number is in place of 8 ... I don't think it would telescope then ?? ... Any idea sir?

Abhinav Raichur - 5 years, 11 months ago

There's only a minimal difference

The answer is 0.2857.

1 solution

Moderator note:

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