+72 Define the following statistic as follows:
\(D(a_1,a_2,\ldots, a_n)\):
Start at \(a_n\). Go back from \(a_n\) until you get to the nearest element of the list smaller than it. Suppose that the first element smaller than it is \(a_k\). We break the list there (right after \(a_k\)).
Then, repeat from \(a_k\)... go back until you get nearest element in the list smaller than \(a_k\).
By the end, you will have split the list a certain number of times, x, creating x+1 sub-lists.
Now, consider the expression:
G(n) = \(\frac{D(1,2,3, \ldots , (n-1) ,n) + D(1,2,3, \ldots , n ,( n-1) , \ldots + D(n , (n-1) , \ldots , 2 , 1)}{n!}\) where the numerator is \(D\) evaluated at all \(n!\) permutations of \(1,2,3,4,\ldots,n\).
For example, for n=3, we have:
\(G(3) = \frac{D(1,2,3)+D(1,3,2)+D(2,1,3)+D(2,3,1)+D(3,1,2)+D(3,2,1)}{3!}\), which can be shown to equal \(\frac{11}{6}.\)
Prove that \(G(n)\) is the \(n\)th harmonic number!!!!!!!!!!!!!!!!!
Thank you so much in advance.
+118687 You asked me to clarify my last answer.
I did that at length.
I am still waiting on a response from you.
