Consider $f\_n(x) = \min\_{|z|=x} \Re \sum\_{j=1}^{n} \frac{z^j}{j}$, a real function of positive variable $x>0$.


I am interested in lower bounds on $f\_n(x)$. Specifically, I ask: what lower bounds can be given on $f\_n(x)$ in the regime where $x=1+O(1/n)$ and $n$ tends to $\infty$?


Trivially, $f\_n(x) \ge - \sum\_{j=1}^{n} \frac{x^j}{j}$ which has order of magnitude $-\log n$ (times a constant) in the aforementioned regime. However, equality cannot be achieved here, since $z^j$ cannot equal to $-x^j$ for both $j=1$ and $j=2$. I don't even know if $f\_n(x)$ is eventually negative.



Asymptotics of truncated logarithm on a cricle
So, we have a function $u\_n(z) = \Re \sum\_{j = 1}^n \frac{z^j}{j}$. As a real part of an analytic function, it is a harmonic function. We are interested in its behaviour on the circle $|z| = x = 1 + \frac{c}{n}$, where $c > 0$ is some constant and we want to estimate the minimal value of $u\_n$ on it. Let me start with the upper bound from my comment:


$u\_n$ is a harmonic function, thus it satisfies the minimum principle in the form $$\min\_{|z| \le x} u\_n(x) = \min\_{|z| = x} u\_n(x)$$ (if you only heard of the maximum principle, then it is it applied to $-u\_n$).


Thus, we have $f\_n(x) \le u\_n(-1)$. On the other hand $u\_n(-1)\to -\log(2)$, so $\limsup f\_n(x) \le -\log(2)$ (in reality, $u\_n(-1) \le -\log(2) + \frac{1}{n}$, so actually $f\_n(x) \le -\log(2) + \frac{1}{n}$).


Now for the lower bound. According to the Lemma 2 from the [link](https://math.dartmouth.edu/%7Ecarlp/pvrev.pdf) provided by Lucia, we have $f\_n(1) \ge -\log(2) - \frac{2}{n}$. For $z$ with $|z| = x$ let us denote $w = \frac{z}{|z|}$. We have


$$u\_n(z) \ge u\_n(w) - |u\_n(z) - u\_n(w)| \ge -\log(2) - \frac{2}{n} - |u\_n(z)-u\_n(w)|.$$


So, if we can give a uniform upper bound on $|u\_n(z)-u\_n(w)|$ then we get a uniform lower bound on $f\_n(x)$. We have
$$u\_n(z) - u\_n(w) = \sum\_{j = 1}^n \frac{z^j-w^j}{j} = \sum\_{j=1}^n w^j \frac{x^j-1}{j}.$$


Since $x = 1 + \frac{c}{n}$ for $j\le n$ we have $1\le x^j\le 1 + \frac{Cj}{n}$ for some $C$ depending on $c$ (something like $C = e^c$ should work). Plugging this in and taking the absolute values (remember that $|w| = 1$) we get
$$|u\_n(z) - u\_n(w)| \le \sum\_{j = 1}^n \frac{C}{n} = C,$$
which proves the desired bound. Note though that the final constant $D$ in $f\_n(x) \ge -D$ depends on $c$ even if we are concerned with only the big values of $n$. I don't know if it can be made approaching $\log(2)$ (it definitely can't be smaller by the minimum principle argument from above).


