I am working on one of my HW assignments $$ \forall n \in \mathbb{Z}, ~ n > 0 ~\rightarrow~ \prod \limits_{i=1}^{n} \left ( \frac{1}{2i~+~1} \cdot \frac{1}{2i~+~2} \right ) ~=~ \frac{1}{(2n~+~2)!} $$ And i am not clear whether it should be proved or disproved. my main concern is base case n=1. $$P_r (1) = \frac {1}{(2*1+1)}+\frac{1}{(2*1+2)} = \frac{1}{12}$$ $$P_l (1) = \frac {1}{(2*1+2)!}= \frac{1}{24}$$ but prove for k+1 works
Induction Hypothesis $$ \prod \limits_{i=1}^{n} \left ( \frac{1}{2i~+~1} \cdot \frac{1}{2i~+~2} \right )~=~ \frac{1}{(2n~+~2)!} \rightarrow \prod \limits_{i=1}^{n+1} \left ( \frac{1}{2i~+~1} \cdot \frac{1}{2i~+~2}\right )~=~ \frac{1}{(2(n+1)~+~2)!} = \frac{1}{(2n+4)!} $$
$$ \prod \limits_{i=1}^{n+1} \left ( \frac{1}{2i~+~1} \cdot \frac{1}{2i~+~2}\right ) = \prod \limits_{i=1}^{n} \left ( \frac{1}{2i~+~1} \cdot \frac{1}{2i~+~2} \right )\cdot \left ( \frac{1}{2(n+1)~+~1} \cdot \frac{1}{2(n+1)~+~2} \right ) $$ by substitution $$ = \frac{1}{(2n~+~2)!}\cdot \left ( \frac{1}{2(n+1)~+~1} \cdot \frac{1}{2(n+1)~+~2} \right ) $$
$$ = \frac{1}{(2n+2)!}\cdot \frac{1}{2n+3} \cdot \frac{1}{2n+4} = \frac{1}{(2n+4)!} $$
Now why my base case isn't working? if it should work for all n>0
$P = \dfrac{1}{3\cdot 5\cdots \cdot (2n+1)}\cdot \dfrac{1}{4\cdot 6 \cdots (2n+2)} = \dfrac{1}{3\cdot 4\cdot 5\cdot 6\cdots \cdot (2n+1)(2n+2)} = \dfrac{1}{\frac{(2n+2)!}{2}}= \dfrac{2}{(2n+2)!}$