How Can Ramanujan's Identity Simplify Summing Powers of Integers?

[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?

 

[CRGreathouse]

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

 

[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...