By monotone convergence (or dominated convergence), let $0< s <1$ we find that as $ s \uparrow 1$ $$ \sum_{k =1 }^\infty \frac{s^k}{k} \to \sum_{k = 1 }^\infty \frac{1}{k} =\infty.$$
My question is, how fast does $\sum_{k =1 }^\infty \frac{s^k}{k}$ go to $\infty$ as $s \uparrow 1$. So, I am looking for something like $\sum_{k =1 }^\infty \frac{s^k}{k} = o(\dots)$.
Note that we could equivalently consider $\sum_{k =1 }^\infty \frac{(1-1/n)^k}{k}$ with $n \to \infty$.
Edit: After the answer given by @Maximlian Janisch a follow up question: Let $(x_k)_k$ be some sequence with $x_k \to 0$ and assume that $$ \sum_{k =1 }^\infty \frac{1}{k} \ x_k = \infty.$$ Then, what can I say about the asymptotic behavior of $$ \sum_{k =1 }^\infty \frac{s^k}{k} \ x_k$$ as $s \uparrow 1$. How much slower than $\log (1-s)$ do we tend to $\infty$ now?
By the Taylor series of the logarithm,
$$\sum_{k =1 }^\infty \frac{s^k}{k}=-\log(1-s)$$ when $s\in[-1,1[$. So it goes to infinity at $1$ "as fast as the logarithm goes to $-\infty$ at $0$".