This comes from the paragraph following equation (27) on page 6 of [this paper](https://arxiv.org/pdf/1506.02438.pdf). 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 $\sum\_{l=0}^\infty\gamma^l\chi(l;s,a)$].
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Given a bound $C$ on $\chi$, those terms can be bounded by $\frac{C\gamma^l}{1-\gamma}$, but why is that negligible?



Given  ∈ (0, 1), why is ˡ negligible for l ≫ 1/(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 $1/e$.


