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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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1  
$S=-\mathrm i\pi$, obviously... –  Did Feb 26 '12 at 14:29
    
@DidierPiau: Great! Please, put this as an answer and I will accept it. Thanks! –  Jon Feb 26 '12 at 14:33
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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. –  sos440 Feb 26 '12 at 14:51

1 Answer 1

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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Raymond, I think Didier is right as $\sum_{n=1}^\infty\frac{2^n}{n}=\Phi(2,1,0)=-i\pi$ as you can check with Mathematica where $\Phi(z,s,a)$ is LerchPhi[z,s,a]. –  Jon Feb 26 '12 at 15:17
    
@Jon: Well Mathematica had to make a choice and $\log(-1)= i \pi\ $ by 'definition' so that $-i \pi\ $ won. This choice seems indeed acceptable. As a physicist you could too consider the symmetry of the situation and accept the mean result $0\ $ (the principal value point of view...) ! –  Raymond Manzoni Feb 26 '12 at 15:25
    
Indeed, I know that this cannot be zero. The reason is that a colleague of mine asked to me the power of a quaternion $\hat i^{\hat i}$, as the case $i^i$ is indeed simple, and, in the effort to give a meaning to this, I met the above sum. –  Jon Feb 26 '12 at 15:38
    
@Jon: Well $i^i= e^{i\log(i)}= e^{i(i\pi/2+2 k\pi i)}= e^{-\pi/2 -2 k\pi}\ \ \ $ so that it admits an infinity of real answers (depending on the branch chosen)! (for the quaternionic 'i' or the classical one). That could be the next move of your colleague... –  Raymond Manzoni Feb 26 '12 at 15:45
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@Jon: my derivation would be (if that's the question) : $\displaystyle \log\left(\hat{\imath}^{\hat{\imath}}\right)= \hat{\imath}\log(\hat{\imath})= \hat{\imath}\log\left(e^{\hat{\imath}\frac{\pi}2+2k\pi\hat{\imath}}\right)= \hat{\imath}\left(\hat{\imath}\frac{\pi}2+2k\pi\hat{\imath}\right)=-\frac{\pi}2-‌​2k\pi\ $ so that $\displaystyle\hat{\imath}^{\hat{\imath}}=e^{-\frac{\pi}2-2k\pi}\ $ It we consider as usual the principal branch (that $\hat{\imath}=e^{\hat{\imath}\frac{\pi}2}\ $ i.e. $k=0\ $) then the answer is simply $e^{-\frac{\pi}2}\ $ –  Raymond Manzoni Feb 26 '12 at 19:41

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