# Sum of all integrers of a certain degree

Is there a general formula for the sum all the terms of the a serie such as:

1^n + 2^n + 3^n ... a^n

?

Werg22 said:
Is there a general formula for the sum all the terms of the a serie such as:
1^n + 2^n + 3^n ... a^n
?
well, as far as i know, the formulas differ depending on the power. I don't think that you can generalize a rhiemann summation like that...although it would defenetly be useful.

i know that sum(n) = n(n-1)/2
sum(n^2)=n(n-1)(n-2)/6

and so on...but i never had to memorize them so i might be wrong about the second one...if there was a way to combine all of them no matter the power, it should be in the precalculus manuals i think.

This looks like one

I was looking for this a few weeks ago and found one here:
http://mathworld.wolfram.com/FaulhabersFormula.html" [Broken]
Also, google: "general power series sum" or variation and you should get quite a few hits

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mathman
There is a standard reference "Table of Integrals, Series, and Products" by Gradshteyn and Ryzhik, published by Academic Press. The answer gets fairly complicated in n. It is a polynomial in degree n+1 without a constant term and sometimes without a linear term. Specifically there are terms in n+1, n, n-1, n-3, n-5, etc.

Another way to do it

I came up with another general way. Borrows a bit from Faulhaber's formula but it's a little simpler and might be a little faster as it uses fewer factorials and no Bernoulli numbers

$$\sum_{j=0}^m j^p \quad = \quad p! \sum_{j=0}^m Q_{pj} R_{pmj}$$

$$Q_{pk} \quad = \quad \left\{ \begin{array}{ll} \sum_{j=0}^k -1^j \frac{ (k - j + 1)^p}{j!(p - j)!} \quad & k < p \\ 1 & k \geq p \end{array}$$

$$R_{pmk} \quad = \quad \prod_{l=1}^p 1 + \frac{m - l}{k}$$

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Robokapp said:
well, as far as i know, the formulas differ depending on the power. I don't think that you can generalize a rhiemann summation like that...although it would defenetly be useful.
i know that sum(n) = n(n-1)/2
sum(n^2)=n(n-1)(n-2)/6
and so on...but i never had to memorize them so i might be wrong about the second one...if there was a way to combine all of them no matter the power, it should be in the precalculus manuals i think.
There's a typo there, it should be
sum(n^2)=n(n+1)(2n+1)/6.
Although n(n-1)(n-2)/6 will always return a possitve integer for n > 2, it does not give you the desired number.

benorin
Homework Helper
Gold Member
Here's one using Lagrange Interpolating Polynomials

Here's one using Lagrange Interpolating Polynomials (eqivalent to simplified polynomial curve fitting here).

$$\sum_{k=1}^{n}k^{p}=\sum_{k=1}^{p+2}\left(\sum_{m=1}^{k}m^{p}\prod_{j=1}^{p+2}\frac{n-j}{k-j}\right)$$

where, in the product, the terms with j=k are omitted.

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Alright. I hope the proof of those sum is not tied to integral as I was looking for such a thing in order to proove the integral!

benorin
Homework Helper
Gold Member
Werg22 said:
Alright. I hope the proof of those sum is not tied to integral as I was looking for such a thing in order to proove the integral!
You can use bivariate induction to prove the above sum.