The Lerch Phi Function is defined as
$$\Phi(-s, \alpha, \nu) = \sum_{k = 0}^{+\infty} \frac{(-s)^k}{(k+\nu)^{\alpha}}$$
Now, in my special case I have $\alpha = 1$, hence it does simply reduce to
$$\Phi(-s, 1, \nu) = \sum_{k = 0}^{+\infty} \frac{(-s)^k}{(k + \nu)}$$
Clearly if I put $\nu = 0$ the series shall be infinite.
But when I compute it with Mathematica (Wolfram Serious Mathematica software, not Online Alpha) it says that
$$\Phi(-s, 1, \nu) = -\ln(1+s)$$
How is this possible?
EDIT
There are also other problems related to that. For example setting $\nu = -1$ and Mathematica says
$$\Phi(-s, 1, -1) = 1+ s\ln(1+s)$$
Which is actually not true, since by its definition (both it starts from $k = 0$ or $k = 1$ there is a point in which the series is infinite).
According to the reference, we have
$$\Phi(-s, \alpha, \nu) = \sum_{k = 0}^{+\infty} \frac{(-s)^k}{(k+\nu)^{\alpha}}$$
for $\Re(\nu)>0$. For $\Re(\nu)\le0$, we have
$$\Phi(-s, \alpha, \nu) = \sum_{k = 0}^{+\infty} \frac{(-s)^k}{[(k+\nu)^2]^{\alpha/2}}$$
where any term $k+\nu=0$ is eliminated.
References:
http://reference.wolfram.com/language/ref/LerchPhi.html