Tuesday, December 4, 2018

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 $$\sum_{i=1}^{n}a_i=\frac{n(a_1+a_n)}{2},$$ so if you want to find the sum of the natural numbers from $1$ to $n$, this equation becomes $$\frac{n^2+n}{2},$$ and the roots of this quadratic are at $n=-1$ and $0$. What I find really interesting is that $$\int_{-1}^0 \frac{n^2+n}{2}dn=-\frac{1}{12}$$ There are a lot of people who claim that the sum of all natural numbers is $-\frac{1}{12}$, 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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