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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

user117911
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    The problem is that the result is false: the left-hand side is twice the right-hand side. The product on the left should run from $i=0$ to $n$, not from $i=1$ to $n$. This doesn’t affect the induction step, but it does make it impossible to establish the base case. – Brian M. Scott Mar 19 '15 at 06:21

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$P = \dfrac{1}{3\cdot 5\cdots \cdot (2n+1)}\cdot \dfrac{1}{4\cdot 6 \cdots (2n+2)} = \dfrac{1}{3\cdot 4\cdot 5\cdot 6\cdots \cdot (2n+1)(2n+2)} = \dfrac{1}{\frac{(2n+2)!}{2}}= \dfrac{2}{(2n+2)!}$

DeepSea
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  • the left part has no factorials in it, only right part. But I understand that $$(2n+1)\cdot (2n+2)$$ will equal to $$ (2n+2)!$$ BUT why the base isn't working out? – user117911 Mar 19 '15 at 06:43
  • its not working because for $n =1, LHS = \dfrac{1}{12} \neq \dfrac{1}{24}= RHS$ – DeepSea Mar 19 '15 at 06:45
  • will disprove it with existential statement $$ \exists n \in \mathbb{Z}, ~ n > 0 ~\rightarrow~ \prod \limits_{i=1}^{n} \left ( \frac{1}{2i~+~1} \cdot \frac{1}{2i~+~2} \right ) ~\neq~ \frac{1}{(2n~+~2)!}$$ – user117911 Mar 19 '15 at 07:11