Product Induction Proof

Theorem
\[(1- \frac{1}{2^2})(1- \frac{1}{3^2})(1- \frac{1}{4^2})...(1- \frac{1}{n^2})=\frac{n+1}{2n}\]

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
\[\begin{equation} \begin{aligned} (1- \frac{1}{2^2})(1- \frac{1}{3^2})...(1- \frac{1}{k^2})(1- \frac{1}{(k+1)^2}) &=\frac{n+1}{2n}(1- \frac{1}{(k+1)^2}) \\ &= (\frac{k^2+2k}{(k+1)^2}(\frac{k+1}{2k})\\ &= \frac{k(k+2)}{(k+1)^2}(\frac{k+1}{2k}) \\ &= \frac{k+2}{2(k+1)}\end{aligned} \end{equation}\]

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