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