Sunday, September 1, 2019

calculus - Integrating the formula for the sum of the first n natural numbers



I was messing around with some math formulas today and came up with a result that I found pretty neat, and I would appreciate it if anyone could explain it to me.



The formula for an infinite arithmetic sum is ni=1ai=n(a1+an)2,

so if you want to find the sum of the natural numbers from 1 to n, this equation becomes n2+n2,
and the roots of this quadratic are at n=1 and 0. What I find really interesting is that 01n2+n2dn=112
There are a lot of people who claim that the sum of all natural numbers is 112, so I was wondering if this result is a complete coincidence or if there's something else to glean from it.


Answer



We have Faulhaber's formula:




nk=1kp=1p+1pj=0(1)j(p+1j)Bjnp+1j, where B1=12



fp(x)=1p+1pj=0(1)j(p+1j)Bjxp+1j



We integrate the RHS from 1 to 0 to get



Ip=01fp(x) dx=(1)pp+1pj=0(p+1j)Bjp+2j



Using the recursive definition of the Bernoulli numbers,




Ip=(1)pBp+1p+1=Bp+1p+1



Using the well known relation Bp=pζ(1p), we get



Ip=ζ(p)



So no coincidence here!


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