\(\prod \limits_{k=2}^{n}\left(1-\frac{1}{k^{2}}\right)=\frac{1}{2}\left(1+\frac{1}{n}\right) \)
mit vollst. Induktion:
n=2 :
\(\prod \limits_{k=2}^{2}\left(1-\frac{1}{k^{2}}\right)=\frac{1}{2}\left(1+\frac{1}{2}\right) \)
<=>\( \left(1-\frac{1}{4}\right)=\frac{1}{2}\left(1+\frac{1}{2}\right) \)
<=>\( \frac{3}{4}=\frac{3}{4} \) Passt also .
Wenn es für n gilt, dann folgt
\(\prod \limits_{k=2}^{n+1}\left(1-\frac{1}{k^{2}}\right) \)
\(\prod \limits_{k=2}^{n}\left(1-\frac{1}{k^{2}}\right) \cdot (1-\frac{1}{(n+1)^{2}}) \)
Ind.annahme einsetzen
\( =\frac{1}{2}\left(1+\frac{1}{n}\right)\cdot (1-\frac{1}{(n+1)^{2}}) \)
\( =\frac{1}{2}\left(1-\frac{1}{(n+1)^{2}}+\frac{1}{n}-\frac{1}{n(n+1)^{2}}\right) \)
\( =\frac{1}{2}\left(1+\frac{1}{n}-\frac{1}{(n+1)^{2}}-\frac{1}{n(n+1)^{2}}\right) \)
\( =\frac{1}{2}\left(1+\frac{1}{n}-\frac{1}{(n+1)^{2}}(1+\frac{1}{n})\right) \)
\( =\frac{1}{2}\left(1+\frac{1}{n}-\frac{1}{(n+1)^{2}}(\frac{n+1}{n})\right) \)
\( =\frac{1}{2}\left(1+\frac{1}{n}-\frac{1}{(n+1)}(\frac{1}{n})\right) \)
\( =\frac{1}{2}\left(1+\frac{1}{n}(1-\frac{1}{(n+1)})\right) \)
\( =\frac{1}{2}\left(1+\frac{1}{n}(\frac{n}{(n+1)})\right) \)
\( =\frac{1}{2}\left(1+\frac{1}{n+1}\right) \) q.e.d.