# Sum of all integrers of a certain degree

1. Oct 14, 2005

### Werg22

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

?

2. Oct 14, 2005

### Robokapp

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.

3. Oct 14, 2005

### ktoz

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

Last edited by a moderator: May 2, 2017
4. Oct 14, 2005

### 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.

5. Oct 16, 2005

### ktoz

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

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

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

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

Last edited: Oct 17, 2005
6. Oct 17, 2005

### Karlsen

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.

7. Oct 20, 2005

### benorin

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.

Last edited: Oct 20, 2005
8. Oct 20, 2005

### Werg22

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!

9. Oct 21, 2005

### benorin

You can use bivariate induction to prove the above sum.