n-3-1-4-n-2-

Question Number 103436 by bramlex last updated on 15/Jul/20

\underset{n = 3}{\prod^{\infty}} (1 - \frac{4}{n^{2}}) = ?

Answered by Worm_Tail last updated on 15/Jul/20

\underset{3}{\prod^{o o}} (1 - \frac{4}{n^{2}}) = a

(1 - \frac{4}{3^{2}}) (1 - \frac{4}{4^{2}}) (1 + \frac{4}{5^{2}}) \dots = a

l n ((1 - \frac{4}{3^{2}}) (1 - \frac{4}{4^{2}}) (1 + \frac{4}{5^{2}}) \dots) = l n (a)

l n (1 - \frac{4}{3^{2}}) + l n (1 - \frac{4}{4^{2}}) + l n (1 + \frac{4}{5^{2}}) \dots = l n (a)

\underset{n = 3}{\sum^{o o}} (l n (1 - \frac{4}{n^{2}})) = l n a

\underset{n = 3}{\sum^{o o}} (l n (\frac{(n + 2) (n - 2)}{n^{2}})) = l n a

\underset{n = 3}{\sum^{o o}} (l n (n + 2) + l n (n - 2) - 2 l n (n)) = l n a

\underset{n = 3}{\sum^{o o}} l n (n + 2) + \underset{n = 3}{\sum^{o o}} l n (n - 2) - 2 \underset{n = 3}{\sum^{o o}} l n (n) = l n a

\underset{n = 3}{\sum^{o o}} l n (n + 2) + (l n (1) + l n (2) + l n (3) + l n (4) + \underset{n = 7}{\sum^{o o}} l n (n - 2)) - 2 (l n (3) + l n (4) + \underset{n = 5}{\sum^{o o}} l n (n)) = l n a

\underset{n = 3}{\sum^{o o}} l n (n + 2) = \underset{n = 7}{\sum^{o o}} l n (n - 2) = \underset{n = 5}{\sum^{o o}} l n (n) = s

s + (l n (1) + l n (2) + l n (3) + l n (4) + s) - 2 (l n (3) + l n (4) + s) = l n a

s + l n (1) + l n (2) + l n (3) + l n (4) + s - 2 l n (3) - 2 l n (4) - 2 s = l n a

s + l n (1) + l n (2) + l n (3) + l n (4) + s - 2 l n (3) - 2 l n (4) - 2 s = l n a

s + s - 2 s + l n (1) + l n (2) - l n (3) - l n (4) = l n (a)

l n (\frac{2}{12}) = l n (a)

\frac{2}{12} = a

a = \frac{1}{6}

Commented by bobhans last updated on 15/Jul/20

w a w . . v i a l o g a r i t h m

Answered by bemath last updated on 15/Jul/20

Commented by bobhans last updated on 15/Jul/20

c o o l l g r a p h i c

Facebook Tweet Pin