Prove-i-2-1-1-i-2-1-2-

Question Number 3298 by prakash jain last updated on 09/Dec/15

Prove

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

Commented by 123456 last updated on 10/Dec/15

1 - \frac{1}{i^{2}} = \frac{i^{2} - 1}{i^{2}}

L = \underset{i = 2}{\prod^{\infty}} 1 - \frac{1}{i^{2}}

\ln L = \ln (\underset{i = 2}{\prod^{\infty}} 1 - \frac{1}{i^{2}})

= \underset{i = 2}{\sum^{\infty}} \ln (1 - \frac{1}{i^{2}})

= \underset{i = 2}{\sum^{\infty}} \ln (i^{2} - 1) - 2 \ln i

Answered by prakash jain last updated on 10/Dec/15

continuing with comment from 123456

\ln L = \underset{i = 2}{\sum^{\infty}} (\ln (i^{2} - 1) - 2 l n (i))

= \underset{i = 2}{\sum^{\infty}} [\ln (i - 1) + \ln (i + 1) - 2 \ln (i)]

This is telescoping sum .

\ln L = - \ln 2

Commented by prakash jain last updated on 10/Dec/15

\ln 1 + \ln 3 - \ln 2 - \ln 2

- 2 \ln 3 + \ln 2 + \ln 4

+ \ln 3 + \ln 5 - 2 \ln 4

So all terms cancel except - \ln 2 .