# Is $\lim_{n \rightarrow \infty} \sum_{k = 1}^\infty \frac{(1-\epsilon(n))^k}{k}$ Finite When $\epsilon(n) \rightarrow 0$?

Consider the sum $$\sum_{k = 1}^\infty \frac{(1-\epsilon)^k}{k}$$ with $\epsilon \in (0,1)$. One can show (by the ratio test, for example) that the sum is finite.

Now, let $\epsilon(n)$ tend to $0$ as $n$ tends toward infinity. What can be said about the limit $$\lim_{n \rightarrow \infty} \sum_{k = 1}^\infty \frac{(1-\epsilon(n))^k}{k}?$$ Does it exist? Is it finite?

The only approach I know would be to find a closed form for $$f(n) = \sum_{k = 1}^\infty \frac{(1-\epsilon(n))^k}{k}$$ in terms of $\epsilon(n)$ and inspect the limit, but I cannot make any progress on this front.

Recall the Taylor series of $\ln(1-x)$ at $0$: $$\ln(1-x)=-\sum_{k=1}^{+\infty}\frac{x^k}{k}\qquad \forall -1\leq x<1.$$
Hence $$\sum_{k=1}^{+\infty}\frac{(1-\epsilon(n))^k}{k}=-\ln \epsilon(n)\longrightarrow +\infty.$$
For the closed form, note that $\dfrac{1}{1-t}=1+t+t^2+t^3+\cdots$ and integrate term by term from $0$ to $x$, where $x$ in your case is $1-\epsilon$.