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:




\sum_{k=1}^n k^p = \frac1{p+1}\sum_{j=0}^p (-1)^j\binom{p+1}jB_jn^{p+1-j},~\mbox{where}~B_1=-\frac12



\implies f_p(x)=\frac1{p+1}\sum_{j=0}^p(-1)^j\binom{p+1}jB_jx^{p+1-j}



We integrate the RHS from -1 to 0 to get



I_p=\int_{-1}^0f_p(x)~\mathrm dx=\frac{(-1)^p}{p+1}\sum_{j=0}^p\binom{p+1}j\frac{B_j}{p+2-j}



Using the recursive definition of the Bernoulli numbers,




I_p=(-1)^p\frac{B_{p+1}}{p+1}=-\frac{B_{p+1}}{p+1}



Using the well known relation B_p=-p\zeta(1-p), we get



I_p=\zeta(-p)



So no coincidence here!


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