\[(1- \frac{1}{2^2})(1- \frac{1}{3^2})(1- \frac{1}{4^2})...(1- \frac{1}{n^2})=\frac{n+1}{2n}\]
Proof
Let
\[P(k)\]
be the statement "\[(1- \frac{1}{2^2})(1- \frac{1}{3^2})(1- \frac{1}{4^2})...(1- \frac{1}{k^2})=\frac{k+1}{2k}\]
".\[(1- \frac{1}{2^2})=\frac{2+1}{2 \times 2}=\frac{3}{4}\]
so \[P(1)\]
is true.Suppose now that
\[P(k)\]
is true for \[n=1,2,...,k\]
. Try and prove true for \[P(k+1)\]
. Then\[\begin{equation} \begin{aligned} (1- \frac{1}{2^2})(1- \frac{1}{3^2})...(1- \frac{1}{k^2})(1- \frac{1}{(k+1)^2}) &=\frac{n+1}{2n}(1- \frac{1}{(k+1)^2}) \\ &= (\frac{k^2+2k}{(k+1)^2}(\frac{k+1}{2k})\\ &= \frac{k(k+2)}{(k+1)^2}(\frac{k+1}{2k}) \\ &= \frac{k+2}{2(k+1)}\end{aligned} \end{equation}\]