After playing around with power series $S_{n, p}=\sum_{k=0}^nk^p$ and finding a formula for each natural power, I tried to also find formulae for imaginary power but couldn't figure out how.

$S_n=\sum_{k=0}^nk^i$, where $i=\sqrt{-1}$

I tried expressing the series using cis and got $S_n=\sum_{k=0}^ncis(ln(n))$ but this didn't help me at all. Any hints or full answer will be appreciated.

  • 1
    $\begingroup$ @DietrichBurde No, the exponent is an imaginary unit. $\endgroup$ – Simply Beautiful Art Jan 29 '17 at 13:14
  • $\begingroup$ Have a look at this and its linked posts. $\endgroup$ – StubbornAtom Jan 29 '17 at 13:40

By the Euler-Maclaurin formula,

$$\sum_{k=1}^nk^i=\zeta(-i)+\frac{n^{i+1}}{i+1}+\frac{n^i}2+\frac{in^{i-1}}{12}-\frac{i(i-1)(i-2)n^{i-3}}{720}+\mathcal O(n^{i-5})$$

this is not an exact formula, but rather an approximation to the problem, where $\zeta(-i)\approx0.0033002237+0.4181554491i$ is the Riemann zeta function. A few values are given to show how it approximates:


Ok, not so good for small values, but it gets better.


As you can see, the accuracy increases as $n\to\infty$.

Alternative forms include


where $\zeta(s,q)$ is the Hurwitz zeta function and $H_n^{(p)}$ is the generalized harmonic number.

  • $\begingroup$ Whats the O at the end of the first equation? $\endgroup$ – Brian010515 Jan 29 '17 at 13:40
  • $\begingroup$ @Brian010515 So the first line goes on forever. It's an infinite series approximation. But everything that comes after what I have is $\mathcal O(n^{i-5})$, meaning as $n\to\infty$, the extra left over stuff behaves like $n^{i-5}$. $\endgroup$ – Simply Beautiful Art Jan 29 '17 at 13:42

It has been pointed out that OP's question is about the finite sum, so my answer does not address the question.

The sum $\sum^\infty_{k=1} k^{-s}$ is absolutely convergent only when $\Re s>1$. So the infinite sum $\sum^\infty k^i$ does not converge. However in the region $\Re s>1$, the sum is analytic, so admits a unique continuation to a maximal domain on the complex domain. This continuation is called the Riemann zeta function, and its value at -i is in some sense related to the infinite sum $\sum^\infty k^i$. That value is approximately 0.0033 + 0.418$i$.

By the way, when $s$ is a natural number, the formula for the sum is called Faulhaber's formula.

$$ \sum_{k=1}^nk^s=\frac{1}{s+1}\sum_{j=0}^s(-1)^j{s+1\choose j}B_jn^{s+1-j} $$ where $B_j$ is the $j$th Bernoulli number. This formula does not make any sense for values of $s$ which are not natural numbers, though.

  • $\begingroup$ Obviously the infinite sum does not converge, but the OP is looking for the finite sum. Please reread the question. $\endgroup$ – Simply Beautiful Art Jan 29 '17 at 13:20
  • $\begingroup$ @SimplyBeautifulArt oops, you're right $\endgroup$ – ziggurism Jan 29 '17 at 13:21
  • $\begingroup$ That's $\sum k^{-s}$. The sum $\sum k^s$ is $\zeta(-s)$ and so converges when $\operatorname{Re}(s)<-1$. $\endgroup$ – Akiva Weinberger Jan 29 '17 at 13:21
  • $\begingroup$ Your formula does not extend to complex $s$. Check the upper bound of the summation. $\endgroup$ – Simply Beautiful Art Jan 29 '17 at 13:29
  • $\begingroup$ @SimplyBeautifulArt good point. $\endgroup$ – ziggurism Jan 29 '17 at 13:31

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