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Infinite product of (r^3 - 8)/(r^3 + 8) from r = 3 | JEE Advanced

JEE Maths question with a full step-by-step solution.

Question

\lim_{n\to\infty}\prod_{r=3}^{n}\frac{r^3-8}{r^3+8} =

\dfrac13

\dfrac25

\dfrac38

\dfrac27

correct

Solution

Step 1:

r^3-8 = \left(r-2\right)\left(r^2+2r+4\right),\qquad r^3+8 = \left(r+2\right)\left(r^2-2r+4\right)

Step 2: Let

\varphi(r) = r^2-2r+4

. Then

\varphi(r+2) = \left(r+2\right)^2-2\left(r+2\right)+4 = r^2+2r+4

\frac{r^3-8}{r^3+8} = \frac{r-2}{r+2}\cdot\frac{\varphi(r+2)}{\varphi(r)}

Both factors telescope. Step 3:

\prod_{r=3}^{n}\frac{r-2}{r+2} = \frac{1\cdot2\cdot3\cdots(n-2)}{5\cdot6\cdot7\cdots(n+2)} = \frac{(n-2)!\,\cdot 4!}{(n+2)!} = \frac{24}{(n+2)(n+1)n(n-1)}

Step 4:

\varphi

is shifted by two, so two factors are left at each end.

\prod_{r=3}^{n}\frac{\varphi(r+2)}{\varphi(r)} = \frac{\varphi(n+1)\varphi(n+2)}{\varphi(3)\varphi(4)}

\varphi(3) = 7,\qquad \varphi(4) = 12,\qquad \varphi(n+1) = n^2+3,\qquad \varphi(n+2) = n^2+2n+4

Step 5:

\prod_{r=3}^{n}\frac{r^3-8}{r^3+8} = \frac{24\left(n^2+3\right)\left(n^2+2n+4\right)}{84\,(n+2)(n+1)n(n-1)}

Numerator and denominator are both quartic in

n

, so

\left(n^2+3\right)\left(n^2+2n+4\right) \sim n^4 ,\qquad (n+2)(n+1)n(n-1) \sim n^4

\lim_{n\to\infty} = \frac{24}{84} = \frac27

Answer: (4)

\dfrac27

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