The following question is obtained from Stanford CS234 Lecture 2 notes, Excercise 3.7
Let $r_i$ denote the reward obtained from transition $s_i\rightarrow s_{i+1}$. Furthermore, the return $G_t$ of a Markov reward process is defined as the discounted sum of rewards starting at time $t$ up to the horizon $H$, and is given by the following mathematical formula $G_t=\sum_{i=t}^{H-1}\gamma^{i-t}r_i, \forall0\leq t\leq H-1$.
Consider an infinite horizon Markov reward process, with bounded rewards, i.e. $\exists M \in (0,\infty), \forall i,|r_i|\leq M $ , and $\gamma<1$. Prove that the return for any episode $G_t$ as defined above converges to a finite limit. The plan is to consider the partial sum $S_N=\sum_{i=t}^{N}\gamma^{i-t}r_i$ and show that {$S_N$} is a Cauchy sequence. I'm trying to show that the sequence is Cauchy, but I can't seem to find a viable upper bound for $|S_n-S_m|$.
Here is my attempt so far:
For any $n,m\geq t$,
\begin{equation} \label{eq1} \begin{split} |S_n-S_m| & = |\sum_{i=t}^{n} \gamma^{i-t}r_i - \sum_{i=t}^{m} \gamma^{i-t}r_i| \\ & \leq |\sum_{i=t}^{n} \gamma^{i-t}r_i| + |\sum_{i=t}^{m} \gamma^{i-t}r_i|\\ & \leq \frac{M}{1-\gamma}(1-\gamma^{n-t+1}) + \frac{M}{1-\gamma}(1-\gamma^{m-t+1}) \\ & = \frac{M}{1-\gamma}(2-\gamma^{n-t+1}-\gamma^{m-t+1}) \\ & = ... \end{split} \end{equation}
The first inequality is justified by triangle inequality.
Second inequality is justified because $|G_t|$ becomes a geometric series.
Would sincerely appreciate any hint on how I could continue from here :)
Follow up from @sudeep5221's comment:
For $n > m \geq t,$
\begin{equation} \label{eq2} \begin{split} |S_n-S_m| & = |\sum_{i=t}^{n} \gamma^{i-t}r_i - \sum_{i=t}^{m} \gamma^{i-t}r_i| \\ & = |\sum_{i=m+1}^{n} \gamma^{i-t}r_i|\\ & \leq \frac{M}{1-\gamma}\gamma^{m+1-t}(1-\gamma^{n-m-1})\\ & \leq \frac{M}{1-\gamma}\gamma^{m+1-t} \\ & < ... \end{split} \end{equation}