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

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions.

A community for students.

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

Mathematics
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

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?
actually those 'simplifications' were a bad idea: $$\prod_{n=2}^N(n-1)^{n^2}\cdot\prod_{n=2}^N(n+1)^{n^2}\cdot\prod_{n=2}^Nn^{-(n+1)^2}\cdot\prod_{n=2}^Nn^{-(n-1)^2}\\\begin{align*}\quad&=\prod_{n=1}^{N-1}n^{(n+1)^2}\cdot\prod_{n=3}^{N+1} n^{(n-1)^2}\cdot\prod_{n=2}^Nn^{-(n+1)^2}\cdot\prod_{n=2}^Nn^{-(n-1)^2}\\&=1^{2^2}\cdot\prod_{n=2}^{N-1}n^{(n+1)^2}\cdot\prod_{n=3}^{N+1} n^{(n-1)^2}\cdot\prod_{n=2}^{N}n^{-(n+1)^2}\cdot\prod_{n=3}^Nn^{-(n-1)^2}\cdot2^{-1^2}\\&=\frac12\cdot\prod_{n=2}^{N-1}n^{(n+1)^2}\cdot(N+1)^{N^2}\cdot\prod_{n=3}^{N} n^{(n-1)^2}\cdot N^{-(N+1)^2}\cdot\prod_{n=2}^{N-1}n^{-(n+1)^2}\cdot\prod_{n=3}^Nn^{-(n-1)^2}\\&=\frac{(N+1)^{N^2}}{2N^{(N+1)^2}}\end{align*}$$
this is a telescoping product and it works best factored

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

Brilliant!
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
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!
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
but yeah, that illustrates the telescoping more clearly :p

Not the answer you are looking for?

Search for more explanations.

Ask your own question