Sum of all integrers of a certain degree

  • Thread starter Werg22
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  • #1
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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

?
 

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  • #2
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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.
 
  • #3
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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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  • #4
mathman
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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.
 
  • #5
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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

[tex]
\sum_{j=0}^m j^p \quad = \quad p! \sum_{j=0}^m Q_{pj} R_{pmj}
[/tex]

[tex]
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}
[/tex]

[tex]
R_{pmk} \quad = \quad \prod_{l=1}^p 1 + \frac{m - l}{k}
[/tex]
 
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  • #6
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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.
 
  • #7
benorin
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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).

[tex]\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)[/tex]

where, in the product, the terms with j=k are omitted.
 
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  • #8
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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!
 
  • #9
benorin
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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.
 

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