find-the-value-of-n-2-n-3-1-n-3-1-and-n-1-1-1-n-2-

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

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) + \dots . + l n (n - 1) - l n (n)

+ l n (2) - l n (3) + l n (3) - l n (4) + \dots + 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} .