What does the p-harmonic series converge to when p = 1 + ε?

Question

In infinitesimal calculus, $\epsilon$ is an infinitesimal number, that is, it is defined to be a number smaller than any real number but greater than $0$.

The p-harmonic series is:

$\displaystyle \sum_{n=1}^{\infty} \frac{1}{n^p}$

It is well known that this series diverges when $p \leq 1$ and converges when $p > 1$.

A lot of teachers like to do examples of this problem with $p$ arbitrarily close to one, but still greater, like $p = 1 + 0.001$.

What I want to know is, what happens when $p$ is infinitesimally close to $1$? In other words what does the series converge to when $p = 1 + \epsilon$?:

$\displaystyle \sum_{n=1}^{\infty} \frac{1}{n^{1 + \epsilon}}$

Is this any different from just taking the limit as $p$ approaches 1?:

$\displaystyle \lim_{p \to 1^+} \sum_{n=1}^{\infty} \frac{1}{n^{p}}$

$1 + \epsilon$ is a number that is greater than $1$, whereas, the limit is getting arbitrarily close to $1$, and presumably that means it gets closer to $1$ than $1 + \epsilon$. However, I don't know what number the series could possible converge to when $p = 1 + \epsilon$.

Answer 1

If $\epsilon\not=0$ is infinitesimal then $\zeta(1+\epsilon)$ will be an infinite hyperreal, since the zeta function has a pole at $z=1$.

Answer 2

In standard mathematics there are no actual infinitesimals. But in any case: $$\sum_{n=1}^\infty \frac{1}{n^{1+\epsilon}} = \zeta(1+\epsilon) = \frac{1}{\epsilon} + \gamma + O(\epsilon) \ \text{as}\ \epsilon \to 0+$$