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I wonder if there is any formula for this sum. $$k^\gamma+(k-1)^\gamma+\cdots+1^\gamma,$$ where $k$ is positive integer and $\gamma\in(0,1)$. And how about $\gamma<0$? Or is there any known asymptotic when $k$ tends to infinity?

Connor
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    The closest result is the Faulhaber's formula, which compute the sum $k^n+(k-1)^n+...+1^n$ as a function of $k$ for all positive integers $n$. – Ángel Valencia Aug 23 '17 at 16:15
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    For any $s > 0, \notin \mathbb{Z}$, we have $$\sum_{n=1}^p n^s - \zeta(-s) \asymp \frac{1}{s+1}\sum_{k=0}^\infty \binom{s+1}{k} (-1)^k B_k p^{s+1-k}$$ This is an exercise in Frank W.J. Olver's book "Asymptotics and Specical Functions".
    Look at $\S8.3$ "Contour integral for the remainder term" for more details.     – achille hui Aug 23 '17 at 16:16
  • Are these not the generalised harmonic numbers in reverse order, with the sign of the exponent reversed - $H_{n,m}=1+\frac{1}{2^m}+\frac{1}{3^m}+...+\frac{1}{n^m}$ ? – James Arathoon Aug 23 '17 at 17:51
  • For the case $0< \gamma <1$, may be this post can help: https://math.stackexchange.com/questions/63986/asymptotic-behaviour-of-sums-of-consecutive-powers/64000#64000 – Davood Aug 23 '17 at 17:52
  • you can bound it by $\frac{(k+1)^{\gamma +1}-1}{\gamma +1}$ and $\frac{k^{\gamma +1}-1}{\gamma +1}$. – Ahmad Aug 23 '17 at 18:28
  • For $k \to \infty$ and $\gamma < -1$, the sum becomes $\zeta(-\gamma)$. But this may not be a case you're interested in. – Michael Seifert Aug 23 '17 at 21:45
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    @Famke : This is NOT a duplicate. The posting to which you linked is about positive integer exponents; this one is about exponents between $0$ and $1. \qquad$ – Michael Hardy Aug 24 '17 at 03:56

2 Answers2

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The sum gives $\sum _{j=1}^k j^\gamma = H_k^{(-\gamma)}$ which is the harmonic number of order $\gamma.$

An excellent approximation I have found is

$$\sum _{j=1}^k j^{\gamma}\approx \left(\frac{\gamma}{12 k}+\frac{k}{\gamma+1} + \frac{1}{2}\right) k^{\gamma}+\frac{1}{2} \gamma \log (2 \pi )-\frac{1}{2}$$

for $\gamma=0.5$ I got the following results. It is an asymptotic approximation, so it works better for large $k$. For negative $\gamma$ the approximation works fine if $|\gamma|<0.5$

$$ \begin{array}{r|r|r} k & H_k^{(-\frac12)}& \textit{approximation}\\ \hline 1 & 1 & 1.1678 \\ 11 & 25.7849 & 25.9523 \\ 21 & 66.2486 & 66.4159 \\ 31 & 117.651 & 117.818 \\ 41 & 178.019 & 178.186 \\ 51 & 246.177 & 246.345 \\ 61 & 321.319 & 321.487 \\ 71 & 402.848 & 403.015 \\ 81 & 490.297 & 490.464 \\ 91 & 583.289 & 583.457 \\ 101 & 681.513 & 681.68 \end{array} $$

Michael Hardy
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Raffaele
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This seems to be studied here:

The generalization of Faulhaber's formula to sums of non-integral powers (McGowna. Parks - Journal of Mathematical Analysis and Applications - v 330, 1, 1 June 2007)

There it's proved that

$$ (a+1) \sum_{n=1}^N n^a = N^\gamma F_a(N) + (a+1) \zeta(-a)+O(N^{-\beta})$$

where $a\ge -1$ and non integer (actually it's also valid for complex $a$ with real part greater than $1$), $\beta = m -a$, $m=\lfloor a+1 \rfloor $, $\gamma = -(m-a)$ and

$$F_a(N)=N^{m+1} + \sum_{k=1}^m (-1)^k {a+1 \choose k}B_k N^{m-k+1}$$

with $B_k$ denoting Bernoulli numbers.

leonbloy
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