Pi/product notation property applications problem

Question

I have recently attempted to simplify this $$ P(n) = \prod_{v=2}^{n} (2 + \frac{2}{v^2 - 1}) $$

I have reached an answer (which is wrong) through the following steps:

rearranging what is inside the brackets

$$\prod_{v=2}^{n} (2 + \frac{2}{v^2 - 1}) = \prod_{v=2}^{n} (\frac{2v^2}{v^2 - 1}) $$ separating the equation into two $$2^n \prod_{v=2}^{n} v^2 \cdot \prod_{v=2}^{n} (\frac{1}{v^2 - 1}) $$ using $ i = v^2$ to get rid of $v^2$ (change the lower bound too) $$2^n \prod_{i=4}^{n} i \cdot \prod_{i=4}^{n} (\frac{1}{i - 1}) $$ using properties to simplify and get rid of pi notation $$ 2^n(n! - 3!) \cdot \frac{1}{(n-1)!-2!}$$ which then simplifies to $$ \frac{2^n(n! - 6)}{(n-1)! - 2} $$

I have since tested if my answer is right by plugging in different values of $n$ and it is definitely not right. I don't understand where I went wrong. Was there a property that I used was invalid? I also wasn't sure about the $i=v^2$ part and just assumed I could do that so maybe that's where I went wrong?

Answer 1

The mistake is in saying "Using $i=v^2$, and then letting $i$ run from $4$ to $n$.

That means that

• in the original expression, $v$ went from $2$ to $n$, taking the values $2, 3, 4,\dots, n$, which means that $v^2$ took the values $4, 9, 16, \dots, n^2$, but
• In the derived expression, $v^2$ takes the values $4, 5, 6, \dots, n$, which is not the same.

Also, a minor mistake you made is that if the product goes from $v=2$ to $v=n$, there are $n-1$ factors, so taking $2$ out of the product should result in $2^{n-1}$, not $2^n$.

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It is generally not a good idea to make a switch to indices that will no longer be consequtive. Instead, I suggest you simplify your equation further, using a couple of things:

1. $\displaystyle\prod_{v=1}^n v^2 = \left(\prod_{v=1}^n v\right)^2$
2. $v^2-1=(v+1)(v-1)$

Answer 2

\begin{align*} \prod_{v=2}^n\frac{2v^2}{v^2-1}&=2^{n-1}\prod_{v=2}^n\frac{v^2}{v^2-1}\\ &=2^{n-1}\color{red}{\prod_{v=2}^n\frac{v}{v-1}} \, \color{blue}{\prod_{v=2}^n\frac{v}{v+1}}\\ &=2^{n-1}\color{red}{\left[\frac{n}{1}\right]} \,\, \color{blue}{\left[\frac{2}{n+1}\right]}\\ &=2^n\frac{n}{n+1} \end{align*}

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Added explanation:

$$\prod_{v=2}^{n}\frac{v}{v-1}=\frac{2}{1} \, \frac{3}{2}\, \frac{4}{3} \cdots \frac{n}{n-1}=\frac{n}{1}$$