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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.

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    @DietrichBurde No, the exponent is an imaginary unit. – 2017-01-29
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    Have a look at [this](http://math.stackexchange.com/questions/18983/why-is-sum-limits-k-1n-km-a-polynomial-with-degree-m1-in-n?noredirect=1&lq=1) and its linked posts. – 2017-01-29

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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:

$$\begin{array}{c|c}n&\sum&\zeta\\\hline1&1&0.49913+0.00010i\\2&1.76924+0.63896i&1.38459+0.31946i\\3&2.22407+1.52954i&1.99665+1.08425i\end{array}$$

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

$$\begin{array}{c|c}n&\sum&\zeta\\\hline10&0.41898+7.84548i&0.37600+7.47349i\\20&-8.93237+11.8340i&-8.43768+11.76128i\\30&-18.82708+10.93402i&-18.34384+11.06237i\end{array}$$

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

Alternative forms include

$$\sum_{k=1}^nk^i=\zeta(-i)-\zeta(-i,n+1)=H_n^{(-i)}$$

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

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    Whats the O at the end of the first equation? – 2017-01-29
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    @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 $n^{i-5}$. – 2017-01-29
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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.

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    Obviously the infinite sum does not converge, but the OP is looking for the finite sum. Please reread the question. – 2017-01-29
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    @SimplyBeautifulArt oops, you're right – 2017-01-29
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    That's $\sum k^{-s}$. The sum $\sum k^s$ is $\zeta(-s)$ and so converges when $\operatorname{Re}(s)<-1$. – 2017-01-29
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    Your formula does not extend to complex $s$. Check the upper bound of the summation. – 2017-01-29
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    @SimplyBeautifulArt good point. – 2017-01-29