Are S(N) and the integral of x^p both divergent in the same way?

[eljose]

let be the divergent series:

1^p+2^p+3^p+.....+N^p=S(N) with p>0 my

question is..how i would prove that this series S would diverge in the form:

S(N)=N^{p+1}/p+1 N--->oo

for the cases P=1,2,3,... i can use their exact sum to prove it but for the general case i can not find any prove..perhaps i should try Euler sum formula ..are the divergent series S(N) equal to the integral:

\int_{0}^{\infty}dxx^{p} they both diverge in the same way.

 

[StatusX]

Try approximating the integral:

\int_0^1 x^p dx

with strips of width 1/N.

 

[shmoe]

You still seem to be having trouble distinguishing between an equality and an asymptotic.

You are apparently trying to show that what you've called S(N) is asymptotic to N^{p+1}/(p+1)? Just compare S(N) with the integrals

\int_0^N x^p dx and \int_1^{N+1} x^p dx.

Euler-Maclaurin summation will work as well, but is not needed for this asymptotic.