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Sum of all integrers of a certain degree

  1. Oct 14, 2005 #1
    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. jcsd
  3. Oct 14, 2005 #2
    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

    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.
  4. Oct 14, 2005 #3
    This looks like one

    I was looking for this a few weeks ago and found one here:
    Also, google: "general power series sum" or variation and you should get quite a few hits
  5. Oct 14, 2005 #4


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    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.
  6. Oct 16, 2005 #5
    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\{
    \sum_{j=0}^k -1^j \frac{ (k - j + 1)^p}{j!(p - j)!} \quad & k < p \\
    1 & k \geq p

    R_{pmk} \quad = \quad \prod_{l=1}^p 1 + \frac{m - l}{k}
    Last edited: Oct 17, 2005
  7. Oct 17, 2005 #6
    There's a typo there, it should be
    Although n(n-1)(n-2)/6 will always return a possitve integer for n > 2, it does not give you the desired number.
  8. Oct 20, 2005 #7


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    Here's one using Lagrange Interpolating Polynomials

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


    where, in the product, the terms with j=k are omitted.
    Last edited: Oct 20, 2005
  9. Oct 20, 2005 #8
    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!
  10. Oct 21, 2005 #9


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    You can use bivariate induction to prove the above sum.
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