Question

Show that
\[\prod_{n=2}^N\frac{(n-1)^{n^2}(n+1)^{n^2}}{n^{(n+1)^2}n^{(n-1)^2}}=\frac{(N+1)^{N^2}}{2N^{(N+1)^2}}\]

Answer 1

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?

Answer 2

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

Answer 3

this is a telescoping product and it works best factored

Answer 4

Brilliant!

Answer 5

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

Answer 6

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!

Answer 7

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

Answer 8

but yeah, that illustrates the telescoping more clearly :p