Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I mean like the formula for $\sum^n_{j=1}j$, or $\sum^n_{j=1}j^2$ extended to things like $\sum^n_{j=1}j^{1.5}$, etc.

share|improve this question
add comment

3 Answers

There is no "simple" formula for an exact evaluation of the sum for non-integers like there is for the integer cases, by which I mean there is no polynomial in $n$ with rational coefficients that yields the desired sum, like there are for the integer cases.

However we know the approximate behaviour of the value of the sum quite well. By the Euler-Maclaurin summation formula (or you could think of the following as just an application of the trapezoidal approximation to integrals) we have:

$$ \sum_{k=1}^n k^s = \frac{n^{s+1} }{s+1} + \frac{n^s}{2} + \mathcal{O}(n^{s-1} ) $$

where $ s>0.$

share|improve this answer
    
How about for $\sum^n_{j=1}j^s$, s>0, not the reciprocal? Thank you for this approximation. –  DumbQuestion Nov 16 '11 at 6:19
    
@DumbQuestion Sorry, I made a foolish typo! See the above edited version. –  Ragib Zaman Nov 16 '11 at 7:41
add comment

The generalized harmonic numbers

$$H_n^{(p)}=\sum_{k=1}^n \frac1{k^p}$$

can be expressed in terms of certain special functions.

One can represent them in terms of Hurwitz zeta functions:

$$H_n^{(p)}=\zeta (p)-\zeta (p,k+1)$$

or equivalently in terms of polygamma functions. See this for details.

share|improve this answer
add comment

The bernoulli-polynomials, or better: their replacement by the use of $\small \zeta() $-terms (call them $\small \zeta()$-polynomials) can be generalized to fractional arguments at the $\small \zeta() $-s . However, while the bernoulli-/zeta-polynomials have finitely many terms at integer exponents due to the vanishing binomial-cofactors, that cofactors do not vanish if the exponent is noninteger, so we get infinite series for fractional exponents.

Define the m'th $\small \zeta()$-polynomial (which sums the powers of exponent m)
$$ z_m(x)=\sum_{c=0}^{\infty} \zeta(-(m-c))\binom{m}{c} x^c $$ . For integer m this reduces to a finite number of terms $$ z_m(x)=\sum_{c=0}^{m} \zeta(-(m-c))\binom{m}{c} x^c - {1 \over m+1} x^{m+1} $$ and is identical to the integrals of the bernoulli-polynomials (the Faulhaber's-and Bernoulli's version of the sum-of-like-powers problem). The requested sum of consecutive powers s(a,b,m) is then $\small s(a,b,m)= z_m(a)- z_m(b+1) $

(This answer crossed just with J.M.'s answer)

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.