A method of evaluating sums in the general case

Ever since I learned that the sum of the first nn integers was n(n+1)2\frac { n\left( n+1 \right) }{ 2 } , I've wondered if there was a way to generally derive an equation for the sum of np{n}^{p} for any positive integer pp.

I actually did manage to find a way to do this using algebra, but it wasn't perfect. My original way was based on the fact that the telescoping sum k=1nkp(k1)p=np\sum _{ k=1 }^{ n }{ { k }^{ p }-{ \left( k-1 \right) }^{ p } } ={ n }^{ p }. This implies that n=1xk=1nkp(k1)p=n=1xnp\sum _{ n=1 }^{ x }{ \sum _{ k=1 }^{ n }{ { k }^{ p }-{ \left( k-1 \right) }^{ p } } } ={ \sum _{ n=1 }^{ x }{ { n }^{ p } } }. There are so many ways to go from here. The idea is that the side with the double sum can be simplified using the binomial formula, and the highest order term in that representation will be ckp1c{k}^{p-1} for some constant cc. Then, you do something a bit clever and use the fact that n=1xk=1nf(k)=n=1x(xn+1)f(n)\sum _{ n=1 }^{ x }{ \sum _{ k=1 }^{ n }{ f\left( k \right) } } =\sum _{ n=1 }^{ x }{ \left( x-n+1 \right) f\left( n \right) } . From here, you are left with a few summations that are doable assuming that you already know the formulas for the sums of nq{n}^{q} for q<pq<p, and one term that requires knowledge of the sum that you are trying to find the general formula for. Luckily, this term is multiplied by a constant (c-c from before, in fact) that is less than 1-1, so you can do some algebra by adding it to both sides then dividing a constant out. I'm interested in the simplest form this can be brought to, as the simplest form I managed to get was

n=1xnp=(1)px(x+1)+p!n=1x(v=0p2(1)pvnv+1[x(pv)+p+1](v+1)!(yv)!)p+1\sum _{ n=1 }^{ x }{ { n }^{ p } } =\frac { -{ \left( -1 \right) }^{ p }x\left( x+1 \right) +p!\sum _{ n=1 }^{ x }{ \left( \sum _{ v=0 }^{ p-2 }{ \frac { { \left( -1 \right) }^{ p-v }{ n }^{ v+1 }\left[ x\left( p-v \right) +p+1 \right] }{ \left( v+1 \right) !(y-v)! } } \right) } }{ p+1 }

which is really an absolute mess. I specifically didn't like the fact that you would need to know all previous exponent sums in order to solve this in the general case (this is to find the sum keeping xx as a variable).

However, I recently realized another way to find the sum that doesn't require the knowledge of the previous sums. Here it is:

This is based off of the fact that, for increasing f(x)f\left( x \right), x=0nf(x)0n+1f(x)dx\sum _{ x=0 }^{ n }{ f\left( x \right) } \le \int _{ 0 }^{ n+1 }{ f\left( x \right)dx } . This is obvious geometrically. Essentially, the area under the curve f(x)f\left( x \right) is greater than the approximated area by left-endpoint Riemann sums using width 11. If we let ak{ a }_{ k } represent the kkth derivative of some function g(x)g\left( x \right), and ak{a}_{k} is monotonic for all kk, and ak=0{ a }_{ k }=0 eventually (for all xx), then the simple observation I mentioned earlier in this paragraph can be used to rewrite x=abg(x)\sum _{ x=a }^{ b }{ g\left( x \right) } as something inside of an integral. The bounds on the summation only directly influence the bounds on the integral, so this method is useful for generalizing sums.

I'll write out either derivation if there is any interest in seeing them, otherwise I'll leave any derivation and further thoughts/application to you.

Note by Austin Antonacci
2 years, 1 month ago

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See Faulhaber's formula.

Jon Haussmann - 2 years, 1 month ago

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Thanks! I'm having fun just messing around with the sum, but I'll look into that.

Austin Antonacci - 2 years, 1 month ago

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