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Let $i=\sqrt{-1}$ the complex imaginary unit, taking $$arg(2)=0$$ for the definition of the summand $2^i$ in $$1^i+2^i+3^i+\cdots (N-1)^i,$$ as $$2^i=\cos\log 2+ i\sin\log 2,$$

see [1].

Question. It is possible to get a closed-form (or the best approximation possible), for an integer $N\geq 1$ $$1+2^i+3^i+\cdots (N-1)^i,$$ where the summands are defined in the same way, taking principal branches of complex argument and complex exponentiation?

Thanks in advance, my goal is start to refresh some easy facts in complex variable, please tell me if there are mistakes in the use of previous definitions.

References:

[1] MathWorld, http://mathworld.wolfram.com/ComplexExponentiation.html http://mathworld.wolfram.com/ComplexArgument.html

2 Answers2

3

I think there is a closed form, notice:

$$1^i+2^i+3^i+\dots+(n-1)^i=\sum_{k=2}^{n}\left(k-1\right)^i=\sum_{k=1}^{n-1}k^i=\text{H}_{n-1}^{(-i)}=\zeta(-i)-\zeta(-i,n)$$

Where $\zeta(s,a)$ is the Hurwitz zeta function, $\zeta(s)$ is the Riemann zeta function and

$\text{H}_{n}^{(r)}$ is the generalized harmoic number.

EDIT:

$$\zeta(-i)=\lim_{s\to0}\left[\sum_{k=1}^{\infty}k^{i-s}\right]\approx 0.0033+0.4182i$$

Jan Eerland
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  • Yesterday too another user wrote this answer, I assume that was right, but can you explain for example what is $\zeta(-i)$? Then you give more useful facts to me and other users. –  Dec 23 '15 at 11:27
  • Very thanks much for previous detail @JanEerlaand –  Dec 23 '15 at 12:17
  • @JuanLópezGonzález You're more than welcome buddy! – Jan Eerland Dec 23 '15 at 12:35
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First of all, you make the assumption $$1^i=1$$When this is not true.$$1^i=e^{\pm2\pi n},n=0,1,2,3,\ldots$$

More generally, I will solve$$1^x+2^x+3^x+4^x+5^x\ldots=\sum_{n=1}^{\infty}n^x$$This has solution obtainable via permutation:$$\sum_{n=2}^{m}n^x+n=\sum_{n=1}^{m-1}n^x+n^m$$$$\sum_{n=1}^{m-1}(n+1)^x+n=\sum_{n=1}^{m-1}n^x+n^m$$Apply Binomial thereom:$$(n+1)^x=\sum_{n=0}^{\infty}\frac{n^{x-n}1^nx!}{(x-n)!n!}$$$$\sum_{n=1}^{m-1}\sum_{i=n}^{\infty}\frac{n^{x-n}1^nx!}{(x-n)!n!}+n=\sum_{n=1}^{m-1}n^x+n^m$$

And now, to be honest with you, I can't proceed from here.

Or perhaps I took it from the wrong method of solving.

Simply Beautiful Art
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  • Very thanks much @SimpleArt , I've read your answer, with respect $1^i$ thanks for clarify my doubt, –  Dec 23 '15 at 07:58
  • OP explicitly states they use principal branches, so $1^i=1$ does hold. – Wojowu Dec 23 '15 at 14:11