TITLE: Is there a formula for sums of consecutive powers, where the powers are non integer?
QUESTION [7 upvotes]: 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.

REPLY [2 votes]: 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.

REPLY [2 votes]: 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)