Bonjour,
[tex]\displaystyle{ln
(\prod\limits_{k = 2}^n \left( {1 - \frac{1} { k^2} } \right))
=\sum\limits_{k = 2}^n {ln \left( {1 - \frac{1} { k^2} } \right) } \\\\
[/tex]
[tex]\displaystyle{ = ln \left( {\left(
{1 - \frac{1}{{2^2 }}} \right) * \left( {1 - \frac{1}{{3^2 }}} \right) *
... * \left( {1 - \frac{1}{{n^2 }}} \right)} \right) }\\\\
[/tex]
[tex]\displaystyle{ = ln\ ( 1 -
\frac{1}{2}) (1 - \frac{1}{3}) * ... * (1 - \frac{1}{n}) * (1 -
\frac{1}{2}) * (1 - \frac{1}{3}) * ... * (1 - \frac{1}{n}) }\\\\
[/tex]
[tex]\displaystyle{ = ln {\left(
{\frac{1}{{\not 2}} * \frac {\not 2 }{\not 3} *...* \frac{n - 1 } {n} }
\right) * \left( \frac{\not 3}{2} *\frac{4}{\not 3} * ... * \frac{n +
1}{n} \right)} \\\\
[/tex]
[tex]\displaystyle{ = ln \left( \dfrac{1}{n} * \dfrac{n + 1}{2} \right)}\\\\
[/tex]
[tex]\displaystyle{ = ln \left( {\dfrac{1 + \dfrac{1}{n}} {2} } \right)} \\\\
[/tex]
[tex]\Rightarrow \sum\limits_{k = 2}^{ + \infty } {ln \left( {1 - \frac{1}{{k^2 }}} \right)} = \ln \dfrac{1}{2}\\\\
[/tex].
[tex]\displaystyle{\prod\limits_{k = 2}^n \left( {1 - \frac{1} { k^2} } \right) =\dfrac{1}{2} }
[/tex]
