The sum of an infinite horizon discounted model is given as follows:
$$R_t = r_{t+1} + \gamma r_{t+2} + \gamma^2r_{t+3} + ... = \sum_{k=0}^\infty \gamma^kr_{t+k+1}.$$
As can be seen, the sum is bounded, but how can I show that there exists a real number $S \in \mathbb{R}$ and $ S < \infty$ such that $R_t < \infty$?
Remark: $r_t < \infty$ for all $t$
My approach is:
For any real parameter $\gamma \in (0,1)$ follows:
$$\lim_{k\to\infty} \sum_{k=0}^k\gamma^k = \frac{1}{1-\gamma}.$$
by simply adding the reward parameter $r_t$ to it, the boundary is given as follows:
$$R_t= \lim_{k\to\infty}\sum_{k=0}^k\left[\gamma^kr_t\right] = \frac{r_t}{1-\gamma}.$$
If I am not mistaken, this should represent the boundary. The value of $R_t$ is now limited by the real number given above.
Is this even the right approach? Any help would help me to understand this totally new topic. Thanks.
We assume that the rewards $r_t$ are bounded and that $0\leq \gamma < 1$ (excluding $\gamma=1$ is important or we cannot guarantee boundedness of the return $R_t$). If the rewards are bounded we can find a number $r_0$ such that $|r_t|\leq r_0$. This inequality is a consequence of the boundedness of the reward $r_t$.
We have
$$|R_t| = |r_{t+1} + \gamma r_{t+2} + \gamma^2r_{t+3} + ...| = |\sum_{k=0}^\infty \gamma^kr_{t+k+1}|\leq \sum_{k=0}^\infty |\gamma^kr_{t+k+1}|= \sum_{k=0}^\infty |\gamma^k||r_{t+k+1}|$$ $$\leq \sum_{k=0}^\infty |\gamma^k|r_0=r_0\sum_{k=0}^\infty |\gamma|^k=r_0\dfrac{1}{1-|\gamma|}.$$
At the last step I used the formula for the infinite geometric series. Hence, we have found that $$|R_t|\leq r_o\dfrac{1}{1-|\gamma|} \implies R_t<\infty.$$
Instead of using $|r_t|\leq r_0$, we could also use the same reasoning by using $r_\text{min}\leq r_t \leq r_\text{max}$ (this follows from the boundedness of $r_t$) to conclude
$$r_\text{min}\dfrac{1}{1-|\gamma|}\leq R_t \leq r_\text{max}\dfrac{1}{1-|\gamma|}.$$