$$ G_n= G_{n-1}+ (\frac{3}{4})^n G_{n-2}, \quad with \quad G_0=1,G_1=1 $$
I am trying to find the limit $ \lim\limits_{n \rightarrow +\infty} G_n$ which converges to $ 5.457177946$.
Or to find an upper bound of $g_n$ that still converges to a constant. Fibonacci sequence does not converges to a constant although it is an upper bound.
I do not need an exact solution, I just a tip or a method name that I could use. I tried to find the general form of the coefficients, but it seems hard to find.
This is not an exact answer but maybe useful. solving your sequence in the general case is difficult and because of this we consider the special case of your sequence as follows $$ G_n^{(k)}= G_{n-1}^{(k)}+ (\frac{3}{4})^k G_{n-2}^{(k)}, \quad with \quad G_0^{(k)}=1,G_1^{(k)}=1 \tag{1} $$ where $k$ is a natural number. By numerical computation(as mentioned) the limit value of your sequence is $5.457177946$. Now the question it is which $k$ is suitable to choose such that the special sequence $(1)$ can estimate the general case of your sequence.
The answer of choosing $k$ is depend on the maximum values that you need for $n$. At first, we can see that the solution of $(1)$, is in the following form $$ G_n^{(k)}=\frac{ \left( -1-\sqrt {1+4\,{3}^{k}{4}^{-k}} \right) \left( -2\,{\frac {{3 }^{k}{4}^{-k}}{1+\sqrt {1+4\,{3}^{k}{4}^{-k}}}} \right) ^{n} }{\sqrt {1+4\,{3}^{k}{4}^{-k}} \left( 1+\sqrt {1+4\,{3}^{k}{4}^{-k}} \right)}+ \frac{ \left( -2\,{\frac {{3}^{k}{4}^{-k}}{1-\sqrt {1+4\,{3}^{k}{4}^{-k}}}} \right) ^{n} }{\sqrt {1+4\,{3}^{k}{4}^{-k}}}\tag{2} $$ We denote the relation $(2)$ by $E(n,k)$. Suppose that the maximum value that you need for $n$ is $1000000$ that means you work with $n\leq 10^6$. In continue, you should find the minimum $k$ such that holds in the following relation $$ E(10^6,k)\leq 5.457177946 \tag{3} $$ the solution of $(3)$ is $k=44$ that means the following sequence $$ G_n^{(44)}= G_{n-1}^{(44)}+ (\frac{3}{4})^{44} G_{n-2}^{(44)}, \quad with \quad G_0^{(44)}=1,G_1^{(44)}=1 \tag{1} $$ is estimate the general case of your sequence for the first $10^6$ elements of the general case with minimum error.
I hope, it be useful.