calculate-n-2-n-3-1-n-3-1-

Question Number 99920 by abdomathmax last updated on 24/Jun/20

calculate \prod_{n = 2}^{\infty} \frac{n^{3} - 1}{n^{3} + 1}

Answered by maths mind last updated on 24/Jun/20

= \underset{n = 2}{\prod^{+ \infty}} \frac{(n - 1) (n^{2} + n + 1)}{(n + 1) (n^{2} - n + 1)}

S = \underset{n = 2}{\prod^{m}} \frac{(n - 1) (n^{2} + n + 1)}{(n + 1) (n^{2} - n + 1)} = \frac{2}{m (m + 1)} . \underset{n = 2}{\prod^{m}} \frac{(n^{2} + n + 1)}{(n^{2} - n + 1)} . . E

n^{2} - n \underset{n = t + 1}{+} 1 = {(t + 1)}^{2} - (t + 1) + 1 = t^{2} + t + 1

E \Leftrightarrow \frac{\underset{n = 2}{\prod^{m}} (n^{2} + n + 1)}{\underset{t = 1}{\prod^{m - 1}} (t^{2} + t + 1)} = \frac{m^{2} + m + 1}{3}

S = \frac{2 (m^{2} + m + 1)}{3 (m^{2} + m)}

S \to \frac{2}{3}

Commented by mathmax by abdo last updated on 24/Jun/20

thank you sir .

Answered by Smail last updated on 24/Jun/20

A = \underset{n = 2}{\prod^{\infty}} \frac{(n - 1) (n^{2} + n + 1)}{(n + 1) (n^{2} - n + 1)}

= \underset{n = 2}{\prod^{\infty}} \frac{n - 1}{n^{2} - n + 1} \underset{n = 2}{\prod^{\infty}} \frac{1}{n + 1} \underset{n = 2}{\prod^{\infty}} (n^{2} + n + 1)

\underset{n = 2}{\prod^{\infty}} \frac{1}{n + 1} = \underset{n = 4}{\prod^{\infty}} \frac{1}{n - 1} = 2 \underset{n = 2}{\prod^{\infty}} \frac{1}{n - 1}

A n d n^{2} + n + 1 = m^{2} - m + 1

\Rightarrow m^{2} - m = n^{2} + n

\Rightarrow {(m - \frac{1}{2})}^{2} = {(n + \frac{1}{2})}^{2} \Rightarrow m = n + 1

S o, \underset{n = 2}{\prod^{\infty}} (n^{2} + n + 1) = \underset{m = 3}{\prod^{\infty}} (m^{2} - m + 1)

= \underset{n = 2}{\prod^{\infty}} (n^{2} - n + 1) \times \frac{1}{3}

A = \underset{n = 2}{\prod^{\infty}} \frac{n - 1}{n^{2} - n + 1} \cdot 2 \underset{n = 2}{\prod^{\infty}} \frac{1}{n - 1} \cdot \frac{1}{3} \underset{n = 2}{\prod^{\infty}} (n^{2} - n + 1)

= \frac{2}{3} \underset{n = 2}{\prod^{\infty}} \frac{(n - 1) (n^{2} - n + 1)}{(n - 1) (n^{2} - n + 1)} = \frac{2}{3}

\underset{n = 2}{\prod^{\infty}} \frac{n^{3} - 1}{n^{3} + 1} = \frac{2}{3}

Answered by mathmax by abdo last updated on 25/Jun/20

let A_{n} = \prod_{k = 2}^{n} \frac{k^{3} - 1}{k^{3} + 1} \Rightarrow A_{n} = \prod_{k = 2}^{n} \frac{k - 1}{k + 1} \prod_{k = 2}^{n} \frac{k^{2} + k + 1}{k^{2} - k + 1}

we have \prod_{k = 2}^{n} \frac{k - 1}{k + 1} = \frac{1}{3} \times \frac{2}{4} \times \frac{3}{5} \times \dots . \frac{n - 1}{n + 1} = \frac{2 (n - 1)!}{(n + 1)!} = \frac{2}{n (n + 1)}

\Rightarrow \ln (A_{n}) = \ln 2 - \ln (n^{2} + n) + \sum_{k = 2}^{n} {\ln (k^{2} + k + 1) - \ln (k^{2} - k + 1)}

= \ln (2) - \ln (n^{2} + n) + \sum_{k = 2}^{n} {x_{k} - x_{k - 1}} with x_{k} = \ln (k^{2} + k + 1)

= \ln 2 - \ln (n^{2} + n) + x_{2} - x_{1} + x_{3} - x_{2} + \dots . + x_{n} - x_{n - 1} = \ln 2 - \ln (n^{2} + n) + x_{n} - x_{1}

= \ln (2) - \ln (n^{2} + n) + \ln (n^{2} + n + 1) - \ln (3) = \ln (\frac{2}{3}) + \ln (\frac{n^{2} + n + 1}{n^{2} + n}) \Rightarrow

\lim_{n \to + \infty} \ln (A_{n}) = \ln (\frac{2}{3}) \Rightarrow \lim_{n \to + \infty} A_{n} = \frac{2}{3}