- #1

zetafunction

- 391

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[tex] \sum_{n=0}^{x}n^{r}= (r+1)^{-1}x^{r+1}+ \zeta (-r) - \sum_{k}B_{2k}\frac{\Gamma (r+1)}{\Gamma (k-2r+2)} [/tex]

does anyone knows how to get this ??

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- Thread starter zetafunction
- Start date

- #1

zetafunction

- 391

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[tex] \sum_{n=0}^{x}n^{r}= (r+1)^{-1}x^{r+1}+ \zeta (-r) - \sum_{k}B_{2k}\frac{\Gamma (r+1)}{\Gamma (k-2r+2)} [/tex]

does anyone knows how to get this ??

- #2

ramsey2879

- 841

- 0

[tex] \sum_{n=0}^{x}n^{r}= (r+1)^{-1}x^{r+1}+ \zeta (-r) - \sum_{k}B_{2k}\frac{\Gamma (r+1)}{\Gamma (k-2r+2)} [/tex]

does anyone knows how to get this ??

To begin with, what is B? and in the sum what is the range of k?

- #3

CRGreathouse

Science Advisor

Homework Helper

- 2,842

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To begin with, what is B? and in the sum what is the range of k?

A Bernoulli number and "0 to infinity", I believe.

- #4

zetafunction

- 391

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oh, excuse me the lack of notation

here B_2k means the 2k-th Bernoulli Number, and yes the summation (is finite) goes to k=0 to k= infinite

i know he used Euler-Maclaurin summation formula but the term [tex] \zeta (-r) [/tex] seems strange to me

here B_2k means the 2k-th Bernoulli Number, and yes the summation (is finite) goes to k=0 to k= infinite

i know he used Euler-Maclaurin summation formula but the term [tex] \zeta (-r) [/tex] seems strange to me

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- #5

damo_clark

- 11

- 0

The "Euler Summation Formula" might give you some insight to why the Bernouli numbers pop up in this formula. The term [tex] \zeta (-r) [/tex] is just [tex] \frac{B_{r+1}}{r+1} [/tex] for a positive integer value of r. Another formula that will compute this sum for a positive integer value of r is :

[tex] f_{r}(x)= \sum_{n=0} ^{x} n^{r} = r\int f_{r-1}dx + xB_{r} [/tex]

where

[tex] f_{1}(x)=\frac{x^2}{2} + \frac{x}{2}[/tex]

I think one of the Bernoulli brothers could calculate [tex] 1^{10}+2^{10}+3^{10} + ... 1000^{10} [/tex] in about fifteen minutes using one of these formulas. Try it using just pen and paper and see how long it takes you...

[tex] f_{r}(x)= \sum_{n=0} ^{x} n^{r} = r\int f_{r-1}dx + xB_{r} [/tex]

where

[tex] f_{1}(x)=\frac{x^2}{2} + \frac{x}{2}[/tex]

I think one of the Bernoulli brothers could calculate [tex] 1^{10}+2^{10}+3^{10} + ... 1000^{10} [/tex] in about fifteen minutes using one of these formulas. Try it using just pen and paper and see how long it takes you...

Last edited:

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