find the value of Π_(n=2) ^∞ ((n^3 −1)/(n^3 +1)) and Π


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Question Number 67534 by mathmax by abdo last updated on 28/Aug/19

$f i n d t h e v a l u e o f \prod_{n = 2}^{\infty} \frac{n^{3} - 1}{n^{3} + 1}$ $a n d \prod_{n = 1}^{\infty} (1 + \frac{1}{n^{2}})$
Commented by ~ À ® @ 237 ~ last updated on 29/Aug/19

$A = \underset{n = 2}{\prod^{\infty}} \frac{n^{3} - 1}{n^{3} + 1} = \underset{n = 2}{\prod^{\infty}} \frac{(n - 1) (n^{2} + n + 1)}{(n + 1) (n^{2} - n + 1)}$ $n^{2} + n + 1 = (n - j) (n - \overline{j}) w i t h j = \frac{- 1}{2} + i \frac{\sqrt{3}}{2}$ $n^{2} - n + 1 = {[(n - 1) + 1]}^{2} - (n - 1) = {(n - 1)}^{2} + 2 (n - 1) + 1 - (n - 1) = p^{2} + p + 1 = (p - j) (p - \overline{j}) w i t h p = n - 1$ $n^{2} - n + 1 = (n - 1 - j) (n - 1 - \overline{j})$ $A_{p} = \underset{n = 2}{\prod^{p}} \frac{n - 1}{n + 1} . \underset{n = 2}{\prod^{p}} \frac{(j - n) (\overline{j} - n)}{[j - (n - 1)] [\overline{j} - (n - 1)]} = \frac{(p - 1)!}{\frac{(p + 1)!}{2!}} . \frac{(j - p) (\overline{j} - p)}{(j - 1) (\overline{j} - 1)} = \frac{2}{(2 - 2 R e (j))} . \frac{(j - p) (\overline{j} - p)}{p^{2}}$ $A = \lim_{n \to \infty} A_{p} = \frac{1}{1 - R e (j)} = \frac{2}{3}$ $K n o w i n g t h a t s i n (π z) = π z \underset{n = 1}{\prod^{\infty}} (1 - \frac{z^{2}}{n^{2}}) l e t t a k e z = i a w i t h a \neq 0$ $s i n (i π a) = (i π a) \underset{n = 1}{\prod^{\infty}} (1 + \frac{a^{2}}{n^{2}})$ $\underset{n = 1}{\prod^{\infty}} (1 + \frac{a^{2}}{n^{2}}) = \frac{s i n (i π a)}{i π a} = \frac{\frac{e^{- π a} - e^{π a}}{2 i}}{i π a} = \frac{s h (π a)}{π a}$
Commented by Abdo msup. last updated on 29/Aug/19
$thank you sir. let A =Π_(n=2) ^∞ ((n^3 −1)/(n^3 +1)) and A_n =Π_(k=2) ^n ((k^3 −1)/(k^(3 ) +1)) ⇒ A_n =Π_(k=2) ^n (((k−1)(k^2 +k+1))/((k+1)(k^2 −x+1))) =Π_(k=2) ^n ((k−1)/(k+1))∐_(k=2) ^n ((k^2 +k+1)/(k^2 −k +1)) ⇒ln(A_n )=Σ_(k=2) ^n ln(((k−1)/(k+1)))+Σ_(k=2) ^n ln(((k^2 +k+1)/(k^2 −k+1))) =Σ_(k=2) ^n {ln(k−1)−ln(k+1)) +Σ_(k=2) ^n { ln(k^2 +k+1)−ln(k^2 −k+1)} =Σ_(k=2) ^n {ln(k−1)−ln(k)}+Σ_(k=2) ^n {ln(k)−ln(k+1)} +Σ_(k=2) ^n (u_k −u_(k−1) ) with u_k =ln(k^2 +k+1) =ln(1)−ln(2)+ln(2)−ln(3)+....+ln(n−1)−ln(n) +ln(2)−ln(3)+ln(3)−ln(4)+...+ln(n)−ln(n+1) +u_n −u_1 =−ln(n)+ln(2)−ln(n+1) +ln(n^2 +n+1)−ln(3) =ln((2/3))+ln(((n^2 +n+1)/(n^2 +n)))→ln((2/3)) ⇒lim_(n→+∞) A_n =(2/3)$
$t h a n k y o u s i r .$ $l e t A = \prod_{n = 2}^{\infty} \frac{n^{3} - 1}{n^{3} + 1} a n d A_{n} = \prod_{k = 2}^{n} \frac{k^{3} - 1}{k^{3} + 1} \Rightarrow$ $A_{n} = \prod_{k = 2}^{n} \frac{(k - 1) (k^{2} + k + 1)}{(k + 1) (k^{2} - x + 1)}$ $= \prod_{k = 2}^{n} \frac{k - 1}{k + 1} ∐_{k = 2}^{n} \frac{k^{2} + k + 1}{k^{2} - k + 1}$ $\Rightarrow l n (A_{n}) = \sum_{k = 2}^{n} l n (\frac{k - 1}{k + 1}) + \sum_{k = 2}^{n} l n (\frac{k^{2} + k + 1}{k^{2} - k + 1})$ $= \sum_{k = 2}^{n} {l n (k - 1) - l n (k + 1))$ $+ \sum_{k = 2}^{n} {l n (k^{2} + k + 1) - l n (k^{2} - k + 1)}$ $= \sum_{k = 2}^{n} {l n (k - 1) - l n (k)} + \sum_{k = 2}^{n} {l n (k) - l n (k + 1)}$ $+ \sum_{k = 2}^{n} (u_{k} - u_{k - 1}) w i t h u_{k} = l n (k^{2} + k + 1)$ $= l n (1) - l n (2) + l n (2) - l n (3) + . . . . + l n (n - 1) - l n (n)$ $+ l n (2) - l n (3) + l n (3) - l n (4) + . . . + l n (n) - l n (n + 1)$ $+ u_{n} - u_{1}$ $= - l n (n) + l n (2) - l n (n + 1) + l n (n^{2} + n + 1) - l n (3)$ $= l n (\frac{2}{3}) + l n (\frac{n^{2} + n + 1}{n^{2} + n}) \to l n (\frac{2}{3}) \Rightarrow {l i m}_{n \to + \infty} A_{n} = \frac{2}{3}$
Commented by Abdo msup. last updated on 29/Aug/19

$w e h a v e s i n (π z) = π z \prod_{n = 1}^{\infty} (1 - \frac{z^{2}}{n^{2}})$ $l e t z = i x (x r e a l) \Rightarrow s i n (i π x) = i π x \prod_{n = 1}^{\infty} (1 + \frac{x^{2}}{n^{2}}) \Rightarrow$ $\prod_{n = 1}^{\infty} (1 + \frac{x^{2}}{n^{2}}) = \frac{s i n (i π x)}{i π x}$ $s i n z = \frac{e^{i z} - e^{- i z}}{2 i} \Rightarrow s i n (i π x) = \frac{e^{- π x} - e^{π x}}{2 i}$ $= - \frac{1}{i} s h (π x) \Rightarrow \prod_{n = 1}^{\infty} (1 + \frac{x^{2}}{n^{2}}) = - \frac{1}{i} \frac{s h (π x)}{i π x}$ $= \frac{s h (π x)}{π x}$ $x = 1 \Rightarrow \prod_{n = 1}^{\infty} (1 + \frac{1}{n^{2}}) = \frac{s h (π)}{π} .$
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