New Math Identity?

[CraigD]

Cindy’s identity is very cool.

 

I don’t have a deep understanding of why it works, but I’m able to generalize it.

 

Using a simplified text notation for summation, here’s Cindy’s equation:

Sum[i=1,n](Sum[j=1,i](j^8)) / Sum[i=1:1:n](Sum[j=1:1:i](j^4)) = (n^4 +4n^3 +n^2 -6n +3)/3

Here’s the general form for these equations:

Sum[i=1,n](Sum[j=1,i](j^(k*m))) / Sum[i=1:1:n](Sum([j=1:1:i](j^k)) = Sum(p=0,m)(c[p]*n^p)

where k and m are positive integers, and c[p] are the coefficients of a degree m polynomial.

 

So there are an infinite family of these identities. Here are a few (lovingly and laboriously calculated by yours truly, so beware of careless errors!):

Sum(i=1,n)(Sum(j=1,i)(j)) / Sum(i=1:1:n)(Sum(j=1:1:i)(j)) = 1 (a trivial case)
Sum(i=1,n)(Sum(j=1,i)(j^2)) / Sum(i=1:1:n)(Sum(j=1:1:i)(j)) = (n +1)/2
Sum(i=1,n)(Sum(j=1,i)(j^3)) / Sum(i=1:1:n)(Sum(j=1:1:i)(j)) = (3n^2 +6n +1)/10
Sum(i=1,n)(Sum(j=1,i)(j^4)) / Sum(i=1:1:n)(Sum(j=1:1:i)(j^2)) = (2n^2 +4n -1)/5
Sum(i=1,n)(Sum(j=1,i)(j^4)) / Sum(i=1:1:n)(Sum(j=1:1:i)(j)) = (2n^3 +6n^2 +3n -1)/10
Sum(i=1,n)(Sum(j=1,i)(j^5)) / Sum(i=1:1:n)(Sum(j=1:1:i)(j)) = (2n^4 +8n^3 +7n^2 -2n -1)/14
Sum(i=1,n)(Sum(j=1,i)(j^6)) / Sum(i=1:1:n)(Sum(j=1:1:i)(j)) = (3n^5 +15n^4 +19n^3 -3n^2 -8n +2)/28

 

There’s something very cool going on here. In particular, these identities don’t work unless k is an integer – that is

Sum[i=1,n](Sum[j=1,i](j^(k*m))) / Sum[i=1:1:n](Sum([j=1:1:i](j^k))

can’t be expressed by as a polynomial of n, eg:

Sum[i=1,n](Sum[j=1,i](j^3)) / Sum[i=1:1:n](Sum([j=1:1:i](j^2))

has no polynomial solution.

[cindy 2005]

To know more similar equations, please see http://www.seriesmathstudy.com. In addion, there are fast convergence series for Ln2 and Ln3 as shown in the attached file (formulaln2ln3.jpg). It is very beautiful series that connect to Pi, Ln2, Ln3, etc... As you know BBP series shows connect to Pi. Here, you can see it connects to Ln2 and Ln3.

[Qfwfq]

That's interesting Cindy! Any idea of how the natural logs come out?

 

Craig, I would need to work your notation into visual form but meantime try a look here:

 

http://mathworld.wolfram.com/PowerSum.html

 

I thought I had posted the link upstream but maybe I hadn't. It is what I used to show why the initial identity is correct.