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I am working on one of my HW assignments $$ \forall n \in \mathbb{Z}, ~ n \geq 2 ~\rightarrow~ \prod \limits_{i=1}^{n} \left ( 1 - \frac{1}{i^2} \right ) ~=~ \frac{n+1}{2n} $$ And i am not clear whether it should be proved or disproved.

my main concern is base case n=2.

$$P_l (2) = \prod \limits_{i=1}^{2} \left ( 1 - \frac{1}{i^2} \right ) = \left ( 1 - \frac{1}{1^2} \right ) \cdot \left ( 1 - \frac{1}{2^2} \right ) = 0 \cdot \frac{3}{4} = 0$$ $$P_r (2) = \frac{2+1}{2 \cdot2} = \frac{3}{4}$$

but prove for k+1 works

Induction Hypothesis $$ \prod \limits_{i=1}^{n} \left ( 1 - \frac{1}{i^2} \right ) ~=~ \frac{n+1}{2n} \rightarrow \prod \limits_{i=1}^{n+1} \left ( 1 - \frac{1}{i^2} \right ) ~=~ \frac{(n+1)+1}{2(n+1)}= \frac{(n+2)}{2(n+1)} $$ $$ \prod \limits_{i=1}^{n+1} \left ( 1 - \frac{1}{i^2} \right ) = \prod \limits_{i=1}^{n} \left ( 1 - \frac {1}{i^2} \right )\cdot \left (1 - \frac {1}{(n+1)^2} \right ) $$ By substitution $$ = \frac{n+1}{2n} \cdot \left (1 - \frac {1}{(n+1)^2} \right ) $$

$$ = \frac{n+1}{2n} \cdot \frac{n(n+2)}{(n+1)^2} $$

$$ = \frac{n \cdot (n+1)\cdot (n+2)}{2n\cdot (n+1)\cdot (n+1)} = \frac{(n+2)}{2(n+1)} $$

Now why my base case isn't working? if it should work for all n>2 Can someone point out what am I doing wrong?

user117911
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  • The result is again false: with the product starting at $i=1$, the left-hand side always has a factor of $1-\frac1{1^2}$, which is $0$, so the left-hand side is always $0$. – Brian M. Scott Mar 19 '15 at 06:35
  • but why (k+1) which is next value, works? – user117911 Mar 19 '15 at 06:40
  • Start the product at 2, not 1, and you will live happily ever after (within 3dB). – marty cohen Mar 19 '15 at 06:55
  • Because *if* $\prod_{i=1}^n\left(1-\frac1{i^2}\right)$ were equal to $\frac{n+1}{2n}$, then $\prod_{i=1}^{n+1}\left(1-\frac1{i^2}\right)$ actually *would* be equal to $\frac{n+2}{2(n+1)}$; but in fact $\prod_{i=1}^n\left(1-\frac1{i^2}\right)$ isn’t equal to $\frac{n+1}{2n}$. – Brian M. Scott Mar 19 '15 at 06:55
  • Sounds like you got some garbage homework or a cruel instructor. – Daniel W. Farlow Mar 19 '15 at 06:58
  • all right then, I'll disprove it with existential statement: $$ \exists n \in \mathbb{Z}, ~ n \geq 2 ~\rightarrow~ \prod \limits_{i=1}^{n} \left ( 1 - \frac{1}{i^2} \right ) \neq \frac{n+1}{2n}$$

    finding counter example will be easy.

     – user117911 Mar 19 '15 at 07:06

1 Answers1

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\begin{align} \prod \limits_{i=1}^{n} \Big(1 - \frac{1}{i^2}\Big)&=\prod \limits_{i=1}^{1} \Big(1 - \frac{1}{i^2}\Big) \times \prod \limits_{i=2}^{n} \Big(1 - \frac{1}{i^2}\Big)\\ &=\prod \limits_{i=1}^{1} \Big(1 - \frac{1}{i^2}\Big) \times \frac{\prod_{i=2}^{n}(i^2-1)}{\prod_{i=2}^{n}i^2}\\ &=\prod \limits_{i=1}^{1} \Big(1 - \frac{1}{i^2}\Big) \times \frac{\prod_{i=2}^{n}(i-1)\prod_{i=2}^{n}(i+1)}{(\prod_{i=2}^{n}i)^2}\\ &=\prod \limits_{i=1}^{1} \Big(1 - \frac{1}{i^2}\Big) \times \frac{(\frac1n\prod_{i=1}^{n}i)\Big(\frac{n+1}{2}\times\prod_{i=1}^{n}i\Big)}{(\prod_{i=1}^{n}i)^2}\\ &=\prod \limits_{i=1}^{1} \Big(1 - \frac{1}{i^2}\Big) \times \frac{n+1}{2n} \end{align}

Math-fun
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