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

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The discussion centers on proving the divergence of the series S(N) = 1^p + 2^p + 3^p + ... + N^p for p > 0, specifically demonstrating that S(N) asymptotically behaves like N^(p+1)/(p+1) as N approaches infinity. The participants suggest using the Euler sum formula and comparing S(N) with integrals such as ∫_0^N x^p dx and ∫_1^{N+1} x^p dx to establish this relationship. Additionally, the Euler-Maclaurin summation is mentioned as a potential tool, although not necessary for this proof.

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eljose
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let be the divergent series:

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

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

[tex]S(N)=N^{p+1}/p+1[/tex] 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:

[tex]\int_{0}^{\infty}dxx^{p}[/tex] they both diverge in the same way.
 
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Try approximating the integral:

[tex]\int_0^1 x^p dx[/tex]

with strips of width 1/N.
 
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 [tex]N^{p+1}/(p+1)[/tex]? Just compare S(N) with the integrals

[tex]\int_0^N x^p dx[/tex] and [tex]\int_1^{N+1} x^p dx[/tex].

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

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