Given a positive integer $n$, consider $f\_n = -\min\_{|z|=1} \Re \sum\_{i>n} \frac{z^i e^{-i/n}}{i}$. What can be said about the growth of $f\_n$? How large can it get?


Taking maximum instead of minimum, the answer is obvious as the maximum is attained at $z=1$ and this becomes a real analysis problem (which in particular implies $f\_n$ is bounded, but perhaps $f\_n$ tends to $0$ at some rate).


An approach suggested by a friend is to express $\sum\_{i>n} t^i/i$ in terms of a Cauchy integral related to $-\ln (1-\zeta t)$ and then perhaps apply a saddle point estimate (uniformly in $|t|=e^{-1/n}$) -- but I wasn't able to carry this out.



Real part of tail of logarithm
The key is the following integral representation:  

\begin{equation}
 \begin{aligned}
s\_n(z)&:= \sum\_{j>n} \frac{z^j e^{-j/n}}j \\ 
 &=\sum\_{j>n} z^j e^{-j/n} \int\_0^\infty du\,e^{-ju} \\ 
 &=\int\_0^\infty du\,\sum\_{j>n} z^j e^{-j/n}e^{-ju} \\ 
 &=\frac{z^{n+1}}{e^{1+1/n}}\,\int\_0^\infty du\,\frac{e^{-(n+1)u}}{1-z e^{-u-1/n}} \\ 
 &=\frac{z^{n+1}}{e^{1+1/n}}\,\int\_0^\infty dv\,\frac{e^{-v-v/n}}{n(1-z e^{-(v+1)/n})}. 
\end{aligned} \tag{1}
\end{equation}
Next, with $n\ge1$, $|z|=1$, and $v>0$,
\begin{equation\*}
 |1-z e^{-(v+1)/n}|\ge1-e^{-(v+1)/n}\ge1-\frac1{1+(v+1)/n}=\frac{v+1}{n+v+1},
\end{equation\*}
whence
\begin{equation\*}
 \Big|\frac{e^{-v-v/n}}{n(1-z e^{-(v+1)/n})}\Big|
 \le2e^{-v}. \tag{2}
\end{equation\*}
Also, for all real $t$ and all real $v>0$,
\begin{equation\*}
 \Big|\frac{e^{-v}}{1+v-it}\Big|\le e^{-v}. \tag{3}
\end{equation\*}


Suppose now that $n\to\infty$ and $z=e^{it/n}$ for some real number $t$ possibly varying with $n$ but staying bounded. Then
\begin{equation\*}
 \frac{e^{-v-v/n}}{n(1-z e^{-(v+1)/n})}=\frac{e^{-v}}{1+v-it}+o(1)
\end{equation\*}
for each real $v>0$.
So, by (1), (2), (3) and dominated convergence,
\begin{align\*}
s\_n(z)&=
\frac{e^{it}}e\,\int\_0^\infty dv\,\frac{e^{-v}}{1+v-it}+o(1)=\Gamma(0,1-i t)+o(1). 
\end{align\*}


On the other hand, if for $z$ with $|z|=1$ it is not true that $z=e^{it/n}$ for some real number $t$ possibly varying with $n$ but staying bounded, then without loss of generality $n(1-z)\to\infty$, whence
\begin{equation\*}
 n|1-z e^{-(v+1)/n}|\ge n|1-z|-(v+1)\to\infty
\end{equation\*}
for each real $v>0$.
So, by (1), (2), and dominated convergence, $|s\_n(z)|\to0$.


So,
\begin{equation\*}
 -f\_n=\min\_{|z|=1} \Re s\_n(z)\to\mu, 
\end{equation\*}
where
\begin{equation\*}
 \mu:=\min\_{t>0}r(t),\quad r(t):=\Re\Gamma(0,1-i t). 
\end{equation\*}
(According to Mathematica's numerics, $\mu=r(t\_\*)=-0.12686\dots$, where $t\_\*=2.0287\dots$.)


