# Shocked by a Ramanujan identity

zetafunction
i do not remember the webpage i watched this but i remember that they said ' IN chapter 1 of his notebook Ramanujan wrote '

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

does anyone knows how to get this ??

ramsey2879
i do not remember the webpage i watched this but i remember that they said ' IN chapter 1 of his notebook Ramanujan wrote '

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

does anyone knows how to get this ??

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

Homework Helper
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.

zetafunction
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 $$\zeta (-r)$$ seems strange to me

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

$$f_{r}(x)= \sum_{n=0} ^{x} n^{r} = r\int f_{r-1}dx + xB_{r}$$

where

$$f_{1}(x)=\frac{x^2}{2} + \frac{x}{2}$$

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

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