The answer is $R=\beta^\frac{-1}{\alpha}$. Let $(i)_n$ be the proposition $$\sum_{1\leq i\leq n}R^{n-i}C_i\leq R^n\cdot M \iff \sum_{1\leq i\leq n}R^{-i}C_i\leq M,$$and $(ii)$ the proposition $$\sum_{1\leq n}\beta^{n}C_{n}^\alpha=\infty.$$We prove a few claims: $\textbf{Claim }1.$ We don't care about $M$. $\textbf{Proof:}$ See that the if a sequence $(C_i)$ satisfies $(i)_n$, and $(ii)$ for a certain $M>0$, then the sequence $\left(\frac{N}{M}C_i\right)$ clearly satisfies $(i)_n$ and $(ii)$ for $M=N$. $\square$ Therefore, we only care whether $\sum_{n\geq1}R^{-n}C_n$ converges or not. Call this property $(i)$. $\textbf{Claim }2.$ If $R=\beta^{\frac{-1}{\alpha}}$, then such a sequence exists. $\textbf{Proof:}$ Just let $$C_nR^{-n}=\frac{1}{n^{\frac{1}{\alpha}}}\iff C_n=\frac{R^n}{n^{\frac{1}{\alpha}}}=\sqrt[\alpha]{\frac{1}{n\beta^n}}.$$Then $(i)$ is just $$\sum_{n\geq1}\frac{1}{n^{\frac{1}{\alpha}}}<\infty,$$which is true since $\alpha\in(0,1)$. Furthermore, $(ii)$ translates into $$\sum_{n\geq1}\beta^nC_n^\alpha=\sum_{n\geq1}\frac{1}{n},$$which diverges. $\square$ $\textbf{Claim }3.$ If $R<\beta^\frac{-1}{\alpha}$, then such a sequence doesn't exist. $\textbf{Proof:}$ Let $R=\lambda\beta^\frac{-1}{\alpha}$, with $0<\lambda<1$. Furthermore, write $$C_n=R^nd_n,$$for some sequence $(d_n)$. Then, $(i)$ implies $$\sum_{n\geq1}R^{-n}C_n<\infty\iff\sum_{n\geq 1}d_n<\infty.$$More specifically, we have $d_n<1$ for $n>N$, for some $N$. But if $(ii)$ is true we must have $$\sum_{n\geq1}\beta^n\cdot \lambda^{\alpha n}\beta^{-n}d_n^\alpha=\infty$$$$\implies\sum_{n\geq1}(d_n\lambda^n)^\alpha=\infty.$$But since we have $d_n\lambda^n<\lambda^n$ for $n>N$, we have $$\underbrace{\sum_{1\leq n\leq N}(d_n\lambda^n)^\alpha}_{C}+\sum_{N<n}(d_n\lambda^n)^\alpha=\infty$$$$\implies C+\sum_{N<n}\lambda^{\alpha n}=\infty,$$which is obviously a contradiction since $\lambda\in(0,1)$. $\square$ Claims $2$ and $3$ conclude the problem. $\blacksquare$

Using the very beautiful proof from @jrsbr of Claim 1, we can conclude the problem as follows Let \[ S_n = \sum_{1\leq i\leq n}R^{-i}C_i \leq M, P_n = \sum_{1\leq i\leq n} \beta ^i C_i^{\alpha}. \] By Holder inequality for exponents $\left ( \frac{1}{\alpha}, \frac{1}{1-\alpha} \right )$ and the sequences \[ \left (\left \{ (R^{\alpha}\cdot \beta)^n \right \}_{n}, \left \{ \left ( \frac{C_n}{R^n} \right )^{\alpha} \right \}_{n} \right ) \] we have \[ P_n \leq S_n^{\alpha} \cdot \left (\sum_{1\leq i\leq n}{(R^{\alpha}\cdot \beta)^{\frac{1}{1-\alpha}n}}\right )^{1-\alpha}. \] As $R, \alpha$ and $\beta$ are constants, if $R^{\alpha}\cdot\beta < 1$, then the right side is limited (by the sum an infinite geometric sequence). So, $\{ C_n \}_{n\geq 0}$ is $R-$inoceronte only if $R^{\alpha}\cdot\beta \geq 1$. To prove that $R = \beta^ {-\frac{1}{\alpha}}$ works, just take $C_n=\frac{1}{(\beta^n n)^{\frac{1}{\alpha}}}$ (and use the mentioned "Claim 1").