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?
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 $2$.