3
$\begingroup$

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?

  • 1
    $S=-\mathrm i\pi$, obviously...2012-02-26
  • 0
    @DidierPiau: Great! Please, put this as an answer and I will accept it. Thanks!2012-02-26
  • 2
    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

1 Answers 1

2

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 $2$.

  • 0
    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].2012-02-26
  • 0
    @Jon: Well Mathematica had to make a choice and $\log(-1)= i \pi\ $ [by 'definition'](http://functions.wolfram.com/ElementaryFunctions/Log/03/02) 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...) !2012-02-26
  • 0
    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.2012-02-26
  • 0
    @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...2012-02-26
  • 0
    In fact, after Didier answer I have got $$\hat i^{\hat i}=e^{-\frac{\pi}{2}}\left(\cosh\frac{\pi}{2}-i\hat i\sinh\frac{\pi}{2}\right)\left(\cosh\left(\frac{\pi}{2}+2k\pi\right)+i\hat i\sinh \left(\frac{\pi}{2}+2k\pi\right)\right)$$ that seems meaningful.2012-02-26
  • 0
    @Jon: I suppose that î is the quaternionic $i$ but what is $i$ in this case?2012-02-26
  • 0
    $i^2=-1$ and $\hat i$ is the quaternion. I have used SU(2) matrices to work out this but I think other means can be used as well.2012-02-26
  • 0
    @Jon: I think this is the point : $\log(î)=\log(e^{î \pi/2+2k\pi î})=î\pi/2+2k\pi î\ \ \ $. The classical $i\ $ doesn't appear at all since $î\ $ plays the same role (I agree that you could add a further $2m\pi i\ $ but that would only confuse things!). For me your $i\ $ should simply be $î\ $ so that your expression simplifies to $\displaystyle e^{-\frac{\pi}2}e^{\frac{\pi}2}e^{-\frac{\pi}2-2k\pi}\ $.2012-02-26
  • 0
    I have worked this out putting $$\hat i=i\sigma_1=i\left(\begin{array}{cc} 0 & 1 \\ 1 & 0 \\ \end{array}\right)$$ and evaluating $(i\sigma_1)^{i\sigma_1}$. Is this wrong?2012-02-26
  • 0
    @Jon: well it is confusing at least (the $i\ $ is part of your conversion from Pauli matrices and should disappear at the end). $\displaystyle i\sigma_1\log(i\sigma_1)=i\sigma_1\log(e^{i\sigma_1\pi/2+2k\pi i\sigma_1})= i\sigma_1(i\sigma_1\pi/2+2k\pi i\sigma_1)\ $ $=-\pi/2-2k\pi\ $ (your problem probably came from the $2k\pi i\ $ without the $\sigma_1\ $, it is much less confusing with quaternions).2012-02-26
  • 0
    It is conventional matter http://en.wikipedia.org/wiki/Quaternion#Matrix_representations . The problem relies on the fact that $e^{i\sigma_1\theta}=\cos\theta+i\sigma_1\sin\theta$. This makes things not so simple.2012-02-26
  • 0
    @Jon: note that the $i\sigma_1\ $ remains as an entity (as in my example). (The quaternions are neat alone : just imagine that a second 'i' exist at right angle with the complex plane, name it 'j' and you'll discover that you need a third one 'k' or your planes will collapse to the standard complex plane). Coming back to your î^î are things clearer now?2012-02-26
  • 0
    Thank you a lot for your helpful explanations. But as you will write down your expression for $\hat i^{\hat i}$?2012-02-26
  • 2
    @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}\ $2012-02-26
  • 0
    Raymond, you have been really helpful. Thank you very much.2012-02-26
  • 0
    @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