The sequence $r_n$ is really a sequence of "averages". So we prove the following statement: If $s_n \to s$, then $\sigma_n \to s$, where
$$
\sigma_n=\dfrac{s_0+s_1+\cdots+s_n}{n+1}
$$
Note that the only difference here is the $n+1$ and that comes from the fact that I begin my sequence at $0$ rather than $1$. Now we prove this statement:
First, observe that
\begin{split}
|\sigma_n-s|&=\left| \frac{s_0+s_1+s_2+\cdots+s_n}{n+1}-s\right| \\
&=\left|\frac{s_0+s_1+\cdots+s_n}{n+1}-\frac{(n+1)s}{n+1}\right| \\
&=\left|\frac{(s_0-s)+(s_1-s)+\cdots+(s_n-s)}{n+1} \right| \\
&=\left| \frac{1}{n+1} \sum_{k=0}^n s_k -s\right| \\
&\leq \frac{1}{n+1} \sum_{k=0}^n |s_n-s|= \frac{1}{n+1} \sum_{k=0}^{N-1} |s_n-s| + \frac{1}{n+1} \sum_{k=N}^n |s_n-s|
\end{split}
Since $\lim_{n\to \infty} s_n=s$, given $\epsilon>0$ there exists a $N \in \mathbb{N}$ such that $|s_n-s|<\frac{\epsilon}{2}$ for $n\geq N$. Moreover, since $s_n$ is a convergent sequence, it is bounded. Then there exists a $M \in \mathbb{N}$ such that $|s_n-s|\leq M$ for all $n$. Then
$$
\frac{1}{n+1} \sum_{k=0}^{N-1} |s_n-s|\leq \frac{NM}{n+1}
$$
Choose $P \in \mathbb{N}$ such that $P>\frac{2NM}{\epsilon}$. Then this implies $P+1>\frac{2NM}{\epsilon}+1>\frac{2NM}{\epsilon}$. But then $\frac{NM}{P+1}<\frac{\epsilon}{2}$. Therefore,
$$
\frac{1}{n+1} \sum_{k=0}^{N-1} |s_n-s| < \frac{\epsilon}{2}
$$
for $n>P$.
Let $\mathcal{J}=\max\{P,N\}$, then
$$
|\sigma_n-s|\leq \frac{1}{n+1} \sum_{k=1}^{N-1} |s_n-s| + \frac{1}{n+1} \sum_{k=N}^n |s_n-s|<\frac{\epsilon}{2}+\frac{(n-N+1)\frac{\epsilon}{2}}{n+1}<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon
$$
for $n>\mathcal{J}$. Therefore, $\sigma_n$ converges to $s$.
Since in your case $s_n \to 1$, $\sigma_n \to 1$. But this idea works more generally as you can see.