anonymous one year ago Show that $\prod_{n=2}^N\frac{(n-1)^{n^2}(n+1)^{n^2}}{n^{(n+1)^2}n^{(n-1)^2}}=\frac{(N+1)^{N^2}}{2N^{(N+1)^2}}$

1. ybarrap

Simplifying the product a bit: $$\prod_{n=2}^N\frac{(n-1)^{n^2}(n+1)^{n^2}}{n^{(n+1)^2}n^{(n-1)^2}}\\ =\prod_{n=2}^N\frac{\left((n-1)(n+1)\right)^{n^2}}{n^{(n+1)^2+(n-1)^2}}\\ =\prod_{n=2}^N\cfrac{(n^2-1)^{n^2}}{n^{2(n^2+1)}}\\ =\prod_{n=2}^N\cfrac{1}{n^2}\left(\frac{n^2-1}{n^2}\right)^{n^2}$$ Using this simplified form, we use Induction to prove: $$=\prod_{n=2}^N\cfrac{1}{n^2}\left(\frac{n^2-1}{n^2}\right)^{n^2}=\frac{(N+1)^{N^2}}{2N^{(N+1)^2}}$$ Basis: $$n=2$$ $$\cfrac{1}{2^2}\left(\frac{2^2-1}{2^2}\right)^{2^2}=\frac{3^4}{2^{10}}\\ =\frac{(2+1)^{2^2}}{2\cdot 2^{(2+1)^2}}=\frac{3^4}{2^{10}}\\$$ Inductive Step: We assume equation holds for $$k$$, then for $$k+1$$ $$=\prod_{n=2}^{k+1}\cfrac{1}{n^2}\left(\frac{n^2-1}{n^2}\right)^{n^2}\\ =\frac{1}{(k+1)^2}\left (\frac{(k+1)^2-1}{(k+1)^2}\right )^{(k+1)^2}\prod_{n=2}^{k}\cfrac{1}{n^2}\left(\frac{n^2-1}{n^2}\right)^{n^2}\\ =\frac{1}{(k+1)^2}\left (\frac{(k+1)^2-1}{(k+1)^2}\right )^{(k+1)^2}\frac{(k+1)^{k^2}}{2k^{(k+1)^2}}\text{ by Induction Hypothesis}\\ =\frac{1}{(k+1)^2}\frac{(k+1)^{k^2}}{(k+1)^{2(k+1)^2}}\times \frac{ \left((k+1)^2-1\right )^{(k+1)^2}}{2k^{k+1)^2}}\\ =\frac{1}{(k+1)^{(2-k^2+2(k+1)^2)}}\times \frac{\left(k^2+2k\right)^{(k+1)^2}}{2k^{(k+1)^2}} \\ =\frac{1}{(k+1)^{((k+1)+1)^2}}\frac{\left( (k+1)+1\right )^{(k+1)^2 }}{2}\\ =\frac{\left( (k+1)+1\right )^{(k+1)^2}}{2(k+1)^{((k+1)+1)^2}}\\$$ QED Does this make sense?

2. anonymous

actually those 'simplifications' were a bad idea: \prod_{n=2}^N(n-1)^{n^2}\cdot\prod_{n=2}^N(n+1)^{n^2}\cdot\prod_{n=2}^Nn^{-(n+1)^2}\cdot\prod_{n=2}^Nn^{-(n-1)^2}\\\begin{align*}\quad&=\prod_{n=1}^{N-1}n^{(n+1)^2}\cdot\prod_{n=3}^{N+1} n^{(n-1)^2}\cdot\prod_{n=2}^Nn^{-(n+1)^2}\cdot\prod_{n=2}^Nn^{-(n-1)^2}\\&=1^{2^2}\cdot\prod_{n=2}^{N-1}n^{(n+1)^2}\cdot\prod_{n=3}^{N+1} n^{(n-1)^2}\cdot\prod_{n=2}^{N}n^{-(n+1)^2}\cdot\prod_{n=3}^Nn^{-(n-1)^2}\cdot2^{-1^2}\\&=\frac12\cdot\prod_{n=2}^{N-1}n^{(n+1)^2}\cdot(N+1)^{N^2}\cdot\prod_{n=3}^{N} n^{(n-1)^2}\cdot N^{-(N+1)^2}\cdot\prod_{n=2}^{N-1}n^{-(n+1)^2}\cdot\prod_{n=3}^Nn^{-(n-1)^2}\\&=\frac{(N+1)^{N^2}}{2N^{(N+1)^2}}\end{align*}

3. anonymous

this is a telescoping product and it works best factored

4. ybarrap

Brilliant!

5. anonymous

Nicely done! This problem came from an interesting question about an infinite product representation of $$\pi$$: $\pi=e^{3/2}\prod_{n=2}^\infty e\left(1-\frac{1}{n^2}\right)^{n^2}$ For those interested, here's the link for the rest of the derivation: http://math.stackexchange.com/a/1346822/170231

6. ybarrap

Ahhh... even simpler! $$\prod_{n=2}^N\frac{(n-1)^{n^2}(n+1)^{n^2}}{n^{(n+1)^2}n^{(n-1)^2}}\\ =\prod_{n=2}^N\frac{(n-1)^{n^2}}{n^{(n+1)^2}}\prod_{n=2}^N\frac{(n+1)^{n^2}}{n^{(n-1)^2}}\\ \text{But, }\\ \prod_{n=2}^N\frac{(n-1)^{n^2}}{n^{(n+1)^2}}=\frac{\cancel{(N-1)^{N^2}}}{N^{(N+1)^2} }\frac{(N-2)^{(N-1)^2}}{\cancel{(N-1)^{N^2} }}\cdots \frac{\cancel{2^{3^2}}}{3^{4^2}}\frac{(2-1)^{2^2}}{\cancel{2^{3^2 }}}=\frac{1}{N^{(N+1)^2}}\\ \text{and }\\ \prod_{n=2}^N\frac{(n+1)^{n^2}}{n^{(n-1)^2}}=\frac{(N+1)^{N^2}}{\cancel{N^{(N-1)^2}}}\frac{\cancel{N^{(N-1)^2}}}{(N-1)^{(N-2)^2}}\cdots \frac{4^{2^2}}{\cancel{3^{2^2}}}\frac{\cancel{3^{2^2}}}{2^{1^2}}=\frac{(N+1)^{N^2}}{2}\\ \text{then}\\ =\prod_{n=2}^N\frac{(n-1)^{n^2}}{n^{(n+1)^2}}\prod_{n=2}^N\frac{(n+1)^{n^2}}{n^{(n-1)^2}}=\frac{1}{N^{(N+1)^2}}\frac{(N+1)^{N^2}}{2}\\$$ Thanks @SithsAndGiggles for this problem!

7. anonymous

eh that's more or less exactly the thing I did, I just separated into four rather than two products so you could see the cancellation more clearly without the risk of losing track of terms

8. anonymous

but yeah, that illustrates the telescoping more clearly :p