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How should I go about evaluating this product? I have not been able to figure out.

$$\lim_{n\to\infty}\prod _{r=3 }^{n }{ \frac { { r }^{ 3 }-{ 8 } }{ { r }^{ 3 }+{ 8 } } } $$

Did
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3 Answers3

9

As an expansion of Did's comment, recognize the difference and the sum of cubes in the general factor to see the product can be rewritten as $$\prod_{r=3}^{\infty} \frac{(r-2)(r^2+2r+4)}{(r+2)(r^2-2r+4)}= \prod_{r=3}^{\infty}\frac{r-2}{r+2}\frac{(r+1)^2+3}{(r-1)^2+3},$$ which makes clear that the $r$-th denominator of the first fraction and the $r+4$-th numerator of it, as well as the $r$-th numerator of the second fraction and the $r+2$-th denominator of it will cancel each other, leaving $$\prod_{r=3}^\infty \frac{r^3-8}{r^3+8}= \frac{1\cdot2\cdot3\cdot4}{7\cdot12}=\frac{2}{7}.$$

Vincenzo Oliva
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We have

$$\begin{align} \prod_{n=3}^{N}\frac{n^3-8}{n^3+8}&=\prod_{n=3}^{N}\frac{(n-2)(n^2+2n+4)}{(n+2)(n^2-2n+4)}\\\\ &=\prod_{n=3}^{N}\frac{(n-2)}{(n+2)}\prod_{n=3}^{N}\frac{(n+1)^2+3}{(n-1)^2+3}\tag 1 \end{align}$$

Now, analyzing the individual partial products we see that the first partial product reduces to

$$\begin{align} \prod_{n=3}^{N}\frac{(n-2)}{(n+2)}&=\frac{1}{5}\frac{2}{6}\frac{3}{7}\frac{4}{8}\frac{5}{9}\cdots\frac{N-6}{N-2}\frac{N-5}{N-1}\frac{N-4}{N}\frac{N-3}{N+1}\frac{N-2}{N+2}\\\\ &=\frac{(1)(2)(3)(4)}{(N-1)(N)(N+1)(N+2)} \tag 2 \end{align}$$

while the second partial product reduces to

$$\begin{align} \prod_{n=3}^{N}\frac{(n+1)^2+3}{(n-1)^2+3}&=\frac{4^2+3}{2^2+3}\frac{5^2+3}{3^2+3}\frac{6^2+3}{4^2+3}\cdots \frac{(N-1)^2+3}{(N-3)^2+3}\frac{(N)^2+3}{(N-2)^2+3}\frac{(N+1)^2+3}{(N-1)^2+3}\\\\ &=\frac{(N)^2+3}{2^2+3}\frac{(N+1)^2+3}{3^2+3}\tag 3 \end{align}$$

Putting $(2)$ and $(3)$ into $(1)$ yields

$$\begin{align} \prod_{n=3}^{N}\frac{n^3-8}{n^3+8}&=\frac27 \frac{(N^2+3)((N+1)^2+3))}{(N+2)(N+1)(N)(N-1)} \end{align}$$

from which we can easily see that

$$\begin{align} \bbox[5px,border:2px solid #C0A000]{\prod_{n=3}^{\infty}\frac{n^3-8}{n^3+8}=\lim_{N\to \infty}\prod_{n=3}^{N}\frac{n^3-8}{n^3+8}=\frac27} \end{align}$$

Mark Viola
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    Isn't this what @Vincenzo explained, plus a colored frame at the end? – Did Jun 21 '15 at 09:15
  • @did It took a while to type this as I was called away from the computer for a while. This is a bit more explicit version and provides a "visual" way of seeing the terms that cancel. It might ptovide for some a primer-type answer. – Mark Viola Jun 21 '15 at 14:30
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HINT: show by induction that $$\prod_{r=3}^k\frac{r^3-8}{r^3+8}=\frac{2(3+k^2)(4+2k+k^2)}{7(k-1)k(k+1)(2+k)}$$

Dr. Sonnhard Graubner
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  • compute the limit, and thx for downvoting – Dr. Sonnhard Graubner Jun 20 '15 at 22:42
  • Is this thread about you it certainly increases my tolerance levels ;). – Chinny84 Jun 20 '15 at 22:42
  • and did you get the limit with the other hints? – Dr. Sonnhard Graubner Jun 20 '15 at 22:48
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    One problem with your answer is that the right hand side of the equality you propose the OP to prove is produced ex nihilo. Why would anyone even consider proving what you tell them? While following your hint will certainly solve the actual question, absolutely no enlightment will ensue. – Mariano Suárez-Álvarez Jun 21 '15 at 00:05
  • It seems that comments taking the Doc to task for the appalling defects of some of their answers are routinely and silently deleted. Is this a mod intervention? Isn't the equivalent of a signpost saying "rubbish" a worthy piece of information to passing by users, regarding answers that are, in effect, rubbish? That the Doc is never addressing these remarks is no excuse, on the contrary. – Did Jun 21 '15 at 09:14