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I am often working with divergent series all around being this the bread and butter for a theoretical physicist. Thanks to the excellent work of Hardy these have lost their mystical Aurea and so, they have found some applications in concrete computations. In these days I have been involved with the following divergent series

$S=\sum_{n=1}^\infty\frac{2^n}{n}$

that is clearly divergent. Wolfram Alpha provides the following

$S_m=\sum_{n=1}^m\frac{2^n}{n}=-i\pi-2^{m+1}\Phi(2,1,m+1)$

being $\Phi(z,s,a)$ the Lerch function (see also Wikipedia). Is there any summation technique in this case?

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    Since logarithm has no natural analytic continuation as a meromorphic function on $\mathbb{C}$, we cannot say $-i\pi$ as a natural choice. Rather, if we understand the summation as the Cauchy principal value of the integral $\int_{0}^{2} \frac{dx}{1-x},$ it should be zero.2012-02-26

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Well $\displaystyle S(x)=\sum_{n=1}^\infty \frac{x^n}n$ so that

$S'(x)=\sum_{n=1}^\infty x^{n-1}=\frac1{1-x}$

and $S(x)=-\log(1-x)+C$

But $S(\frac 12)=\log(2)$ so that $C=0$

and $S(2)$ 'could be' $-\log(1-2)= -\log(e^{\pi i})= -\pi i$ or $-\log(e^{-\pi i})= \pi i\cdots$.

Both choices seem ok by analytic expansion (or none of them since it is at the border! :-)).

Here is a picture of $\mathrm{Im}(-\log\left(1-(x+iy))\right)$ with two continuous paths possible from $z=\frac 12$ to $2$.

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    @Jon: A last word about this : [google](http://www.google.com/#hl=en&gs_nf=1&cp=16&gs_id=24&xhr=t&q=i+to+the+power+i) has some other answers but my favorite now is this one from [Wikipedia](http://en.wikipedia.org/wiki/Imaginary_unit#i_raised_to_the_i_power) : $\displaystyle \hat{\imath}^{\hat{\imath}}= \left(e^{\hat{\imath}\frac{\pi}2+2k\pi\hat{\imath}}\right)^{\hat{\imath}}= e^{-\frac{\pi}2-2k\pi}\ $2012-02-26