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

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 $p = 1 + \epsilon$.

## 2 Answers

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

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

• You are equivocating on the meaning of the term "standard" since you are clearly not referring to it in its technical meaning (referring to Robinson's framework for analysis with infinitesimals) but rather as a generic term of approval. I object to this kind of obfuscation. -1 – Mikhail Katz Jan 25 '18 at 9:51