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I am curious about whether a closed form expression of $$\sum_{k=1}^{n}k^\alpha $$ for $\alpha \in \mathbb{R}$ exists in terms of special functions. Clearly for the natural number case we have the Faulhaber formulae, and when $\alpha = -1$ it is known that we can make use of the digamma function $\psi(x) := \frac{d}{dx}\log(\Gamma(x))$ to create an analogue of $\int\frac{1}{x} dx = \log(x)$ allowing us to sum expressions of the form:

$$\sum_{k=1}^{n}\frac{1}{k+p} = \psi(n+p+1) - \psi(1+p) = \psi(n+p+1) + \gamma - H_p$$

Where $\gamma$ is the Euler-Mascheroni constant and $H_p$ the $p$'th Harmonic number (the formula above is prettier when expressed using the methods of discrete calculus, but I don't want this question to become too niche, see Concrete Mathematics or this nice pdf for the curious).

Is there a slightly more general idea like this? I am more interested in $\sum\limits_{k=1}^{n}\sqrt{k}$, $\sum\limits_{k=1}^{n}\frac{1}{k^m}, m \in \mathbb{N}$ rather than $\sum\limits_{k=1}^{n}\frac{1}{(k+p)^m}$ or arbitrary reals, but I suppose if such a generalisation exists it probably goes the full mile.

I have seen some interesting identities involving polygamma functions and the Hurwitz Zeta, but not quite to the level (at least from what I can glance) of giving a closed form.

Theo Diamantakis
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    You may indeed rewrite the partial zeta sum using the Hurwitz zeta function :$$\zeta(s,n) := \sum_{k=0}^\infty \frac{1}{(n+k)^s}$$ so that :$$\sum_{k=1}^{n}k^\alpha=\zeta(-\alpha,1)-\zeta(-\alpha,n+1)$$ See too this and this threads. – Raymond Manzoni Mar 15 '18 at 23:26
  • Thanks, I found out the generalization of the polygamma function $\displaystyle \psi^{(m)} (z) = (-1)^{m+1} \int_{0}^{\infty} \frac{t^m e^{-zt}}{1-e^{-t}}dt$ provides a suitable representation which can then be written out as a Hurwitz zeta function. Those links are definitely helpful for anyone else who sees this. – Theo Diamantakis Mar 15 '18 at 23:46
  • Harmonic numbers are basically digamma function, differ from it only by a constant: $\psi(x)=\gamma+H(x+1)$ – Anixx Jan 20 '21 at 01:17

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$s=\Sigma^n_{k=1}k^a=1 ^a+2^a+3^a+ . . . k^a$

$e-1=\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+ . . .+\frac{1}{k!}+ ...$

Term by term multiplication of two series gives:

$s . (e-1)=\Sigma^n_{k=1}k^a.\Sigma^n_{k=1}\frac{1}{k!}=\Sigma^n_{k=1}k^a.\frac{1}{k!}=\frac{1^a}{1!}+\frac{2^a}{2!}+\frac{3^a}{3!}+ . . .$

Also:

$e^{e^x}=e[1+x+x^2+(5/6)x^3 +(5/8)x^4+ . . .(c_a) x^a]$

We can see that the coefficients of $x^a$, i.e $(e.c_a)$ in this expansion is:

$e.c_a=\frac{1}{a!}.[\frac{1^a}{1!}+\frac{2^a}{2!}+\frac{3^a}{3!}+ . . .]$

Therefore :

$e. c_a.a!=s(e-1)$

⇒ $s=\Sigma^n_{k=1}k^a=\frac{a!.e.c_a}{e-1}$

sirous
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  • Is there a name for this term by term multiplication of series? Can it be expressed in a concrete way like the Cauchy product? Also how did you obtain $\sum_{k=0}^{\infty}\frac{e^{xk}}{k!} = e[1 + x + x^2 + (5/6)x^3 + \ldots]$. If you have any additional resources or got this from a book so I could read more I would be extremely grateful. Thanks. – Theo Diamantakis Mar 07 '18 at 17:25
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    @TheoDiamantakis, thanks for your attention. If you find this book: "Classical Algebra", G. Praha, Books and allied (p) Ltd. (1999), go to pages 64 to 70, you will see all these formulas. – sirous Mar 07 '18 at 18:51
  • Many thanks for this reference, then I am going to follow this post. –  Mar 08 '18 at 14:24
  • @sirous If possible do you have the ISBN, a link or the Author's full name? I have scoured several search engines and am almost convinced the man or his book does not exist. – Theo Diamantakis Mar 08 '18 at 21:28
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    @TheoDiamantakis, full name of author is Dr. Gunadhar Paria, adress of publisher: 8/1 Chintamoni Das Lane, Calcutta 700 009 (India). Fax:(+91-33)241-3852, Email: [email protected] – sirous Mar 09 '18 at 03:52