Given 𝛾 ∈ (0, 1), why is 𝛾ˡ negligible for l ≫ 1/(1-𝛾)? [closed]

Question

This comes from the paragraph following equation (27) on page 6 of this paper. It's not crucial to the argument — any such bound will do — but it's not clear to me why this particular bound is appropriate.

Using a discount $\gamma < 1$ corresponds to dropping the terms with $l ≫ 1/(1 − γ)$ [in the equation $\sum_{l=0}^\infty\gamma^l\chi(l;s,a)$].

Given a bound $C$ on $\chi$, those terms can be bounded by $\frac{C\gamma^l}{1-\gamma}$, but why is that negligible?

Answer 1

For $\gamma$ approaching 0, $\gamma^{1/(1-\gamma)}$ approaches $\gamma$.

For $\gamma$ approaching 1, $\gamma^{1/(1-\gamma)}$ approaches $1/e$. If $l$ is much larger than $1/(1-\gamma)$, then $\gamma^l$ is much smaller than $1/e$.