# Proving that (4-2/1)(4-2/2)…(4-2/n) in an integer.

We have to prove that $(4-2/1)(4-2/2)...(4-2/n)$ is an integer for $n\in\mathbb{N}$. Can we do this by induction?

We prove for $n = 1$, which is trivial as $(4-2/1) = 2$ which is clearly an integer.

next we assume that the statement is true for some integer $n$ i.e. that $(4-2/1)(4-2/2)...(4-2/n) = p$ where p is an integer, and we prove for $n+1$.

So we need to show that $(4-2/1)(4-2/2)...(4-2/n)(4-2/(n+1))$ is also an integer. Which is the same as saying $p\times(4-\frac{2}{n+1})$ is an integer.

We can rewrite $p$ as $\dfrac{2^n(1\times3\times5\times...\times(2n-1))}{n!}$ so essentially all we need to prove is that $n+1$ divides the numerator of $p$. So assume that $n$ is even, which implies that $n+1$ is odd and certainly less than $2n-1$ so therefore it will divide one of the odd numbers in the numerator of $p$. If $n$ is odd, then $n+1$ is even and will not have more factors of $2$ than $2^n$. So if the prime factorization of $n+1$ contains $k$ factors of 2, $k<n$ or $k=n$, implying that $2^k$ divides $2^n$. Also the prime factorisation of $n+1$ will not have have an odd factor that is greater than $2n-1$ as it already has a factor of 2. Therefore $n+1$ divides the numerator of p.

Can we do this? It seems like it is not entirely correct! I may have made a big mistake in my reasoning.

Any help is appreciated.

We can write $p$ on this form

$$p=\frac{2^n(1\times 3\times 5\times \cdots\times(2n-1))}{n!}=\frac{(2n)!}{n!n!}={2n\choose n}\leftarrow \text{integer}$$

$r$th term $\displaystyle T_r=4-\frac2r=2\left(\frac{2r-1}r\right)$

So, $$\prod_{r=1}^n2^n\frac{\prod_{r=1}^n(2r-1)}{n!}$$

$$=\frac1{n!}\prod_{r=1}^{n-1}\frac{(2r-1)2r}r(2n-1)$$

$$=\frac{(2n-2)!\cdot(2n-1)}{n!\cdot(n-1)!}=\binom{2n-1}n$$

[Cannot seem to add a comment (re request above on the induction method), so posting a new answer in response]

Assume result is true for $n$ terms, i.e. product is $$2n\choose n$$ which is an integer.

Then, for $n+1$ terms, product becomes \begin{align} {2n\choose n}\left( 4-\frac 2{n+1}\right) &=\frac{4n+2}{n+1} {2n\choose n} \\ &=2\cdot \frac{2n+1}{n+1}\cdot {2n\choose n} \\ &=\frac {2n+2}{n+1}\cdot \frac{2n+1}{n+1}\cdot\frac{2n(2n-1)...(n+2)(n+1)}{1\cdot2\cdot ... n} \\ &=\frac {(2n+2)(2n+1)(2n)...(n+2)}{1\cdot 2\cdot ... n(n+1)}\\ &={{2n+2}\choose{n+1}}\\ &={{2(n+1)}\choose{n+1}}\\ \end{align} i.e. also true for $n+1$.

It can easily be shown that the product is true for $n=1$.

Hence, by induction, the proposition is true for all integer $n$.

The product is $$2 {{2n-1}\choose {n-1}}$$ which is an integer for integer $n$.

This is also equal to $$2n \choose n$$ as pointed out by another member earlier.