Question

The value of \(\lim_{n \rightarrow \infty} \prod_{r=3}^{n} \frac{r^3 - 8}{r^3 + 8}\) equals to:

• A. \(2/7\)
• B. \(3/7\)
• C. \(4/7\)
• D. \(6/7\)

Hint:

For infinite products involving cubic expressions, first factorize completely and then look for telescoping cancellation patterns.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
We need to evaluate: \[ \lim_{n\to\infty}\prod_{r=3}^{n}\frac{r^3-8}{r^3+8} \]

Step 2: Key Formula or Approach:
Factorize using: \[ a^3-b^3=(a-b)(a^2+ab+b^2) \] and \[ a^3+b^3=(a+b)(a^2-ab+b^2) \] Thus: \[ r^3-8=(r-2)(r^2+2r+4) \] \[ r^3+8=(r+2)(r^2-2r+4) \] Observe: \[ r^2+2r+4=(r+1)^2+3 \] \[ r^2-2r+4=(r-1)^2+3 \] This creates telescoping cancellation.

Step 3: Detailed Explanation:
Write the product: \[ P_n=\prod_{r=3}^{n}\frac{(r-2)(r^2+2r+4)}{(r+2)(r^2-2r+4)} \] Separate terms: \[ P_n= \left(\prod_{r=3}^{n}\frac{r-2}{r+2}\right) \left(\prod_{r=3}^{n}\frac{r^2+2r+4}{r^2-2r+4}\right) \] Now: \[ \prod_{r=3}^{n}\frac{r-2}{r+2} = \frac{1\cdot2\cdot3\cdots(n-2)} {5\cdot6\cdot7\cdots(n+2)} \] \[ = \frac{4! \, n!}{2!(n+2)!} = \frac{12}{(n+1)(n+2)} \] Also, \[ \frac{r^2+2r+4}{r^2-2r+4} = \frac{(r+1)^2+3}{(r-1)^2+3} \] Hence: \[ \prod_{r=3}^{n}\frac{r^2+2r+4}{r^2-2r+4} = \frac{(4^2+3)(5^2+3)\cdots((n+1)^2+3)} {(2^2+3)(3^2+3)\cdots((n-1)^2+3)} \] Most terms cancel: \[ = \frac{(n^2+2n+4)(n^2+4n+7)} {7\cdot12} \] Therefore: \[ P_n= \frac{12}{(n+1)(n+2)} \cdot \frac{(n^2+2n+4)(n^2+4n+7)} {84} \] \[ = \frac{(n^2+2n+4)(n^2+4n+7)} {7(n+1)(n+2)} \] Taking limit as \(n\to\infty\): \[ \lim_{n\to\infty}P_n=\frac{4}{7} \]

Step 4: Final Answer:
The correct option is: \[ \boxed{\text{(C) }\frac{4}{7}} \]