I have this situation: supposing that $\Omega$, is an open, bounded subset of $\mathbb{R}^N$, $N>2$. Given $t_0>0$ and $\overline{n}>0$, let's consider $\overline{t}=t_0+1$, $m>\overline{t}$ and $n>\max\{\overline{t}, \overline{n}\}$. We are also considering $\gamma>1$ and $2^{*}:=2N/(N-2)>\gamma+2$ and $p$, $q$ such that
- $\gamma+2<p<\frac{2^*}{2}(\gamma+2)$
- $\gamma+2<q<\min\{p,2^*\}$.
Taking $u_{m,n}$ as a weak solution of some Dirichlet problem. Assuming that $u_{m,n}\in L^{r}(\Omega)$ for some $r>1$ and choosing $r-q>0$, I proved that $u_{m,n}\in L^{\frac{r-1+2}{2}2^*}(\Omega)$, with
\begin{equation*}\scriptsize |u_{m,n}|_{L^{\frac{r-q+2}{2}2^*}(\Omega)}\leq \left(\frac{(r-q+2)^2}{4\alpha^2 (r-q+1)}\right)^{\frac{1}{r-q+2}}n^{\frac{p-q}{r-q+2}}|u_{m,n}|_{L^r(\Omega)}^{\frac{r}{r-q+2}}\tag{1}\label{1} \end{equation*}
where, $\alpha>0$. Now, supposing that $u_{m,n}\in H_{0}^{1}(\Omega)$ by Sobolev embedding we also have that, $u_{m,n}\in L^{2^*}(\Omega)$. Until here everything was ok. The problem starts after this situation: I had used Moser iteration (I'm not sure if this is the best solution). So, let me explain: Choosing $r=r_0=2^*$, I deduced that $u_{m,n}\in L^{\frac{r_0-q+2}{2}2^*}(\Omega)$. Now, choosing $r=r_1=\frac{r_0-q+2}{2}2^*$ I obtained that $u_{m,n}\in L^{\frac{r_1-q+2}{2}2^*}(\Omega)$. Iterating this process, and defining by induction $$r_k:=\frac{2^*}{2}\left(r_{k-1}+2-q\right).$$ I obtained by \eqref{1} that
\begin{equation*}\scriptsize|u_{m,n}|_{{r_k}}\leq \left[\left(\frac{(r_{k-1}-q+2)^2}{4\alpha^2(r_{k-1}-q+1)}\right)n^{(p-q)}\right]^{\frac{1}{r_k}\displaystyle\sum_{i=1}^{k}\left(\frac{2^*}{2}\right)^i}|u_{m,n}|_{{2^*}}^{\left(\frac{2^*}{2}\right)^k\frac{r_0}{r_k}}.\tag{2}\label{2} \end{equation*}
I'm struggling right here. Supposing $r_k\rightarrow \infty$, how can I be sure that both exponents in expression \eqref{2} (involving $r_k$) are not blowing up? I mean, is there a way to prove that the numerator in both exponents goes to infinity slower than the denominator? Another question is how much is $\sum_{i=1}^{k}\left(\frac{2^*}{2}\right)^i$ (step-by-step)? I know that is a geometric series with a ratio greater than 1, but I was not able to write in a way that helps me to answer the first doubt.
