# Find the limit of a recursively defined sequence.

$$F_k=\begin{cases} 4G_{k-1},\ k>1\\ 0,\ k=1 \end{cases} \\ G_k=\begin{cases} 4^{k-1}-2G_{k-1},\ k>1\\ 1,\ k=1 \end{cases}$$ Find $$\lim_{k\rightarrow\infty}\frac{F_k}{4^k}$$

Here is what I did: I substituted $$F_k$$ with $$G_k$$ in the limit, i.e. $$\lim_{k\rightarrow\infty}\frac{F_k}{4^k}= \lim_{k\rightarrow\infty}\frac{G_{k-1}}{4^{k-1}}$$ But I have no idea how to find such a limit. If anyone could give me some clue, I would be grateful.

Hint. If the limit exists and it is equal to $$L$$ then by letting $$k\to \infty$$ on both sides we find $$L\leftarrow\frac{G_k}{4^k}=\frac{4^{k-1}-2G_{k-1}}{4^k}\rightarrow \frac{1}{4}-\frac{L}{2}\implies L=\frac{1}{6}.$$
P.S. Note that the linear recurrence $$G_k+2G_{k-1}=4^{k-1}$$ with $$G_1=1$$ can be easily solved: $$G_k=A(-2)^k+B4^k$$ where $$A,B$$ are real numbers to be determined.
• But how can I prove that L exists if $\frac{G_k}{4^k}$ is not monotonous and bounded at the same time? Oct 30 '19 at 10:34
• $a_k=\frac{G_k}{4^k}$ is bounded but not monotone. Consider the subsequence $a_{2k}$ and $a_{2k+1}$. Or see my P.S. and find $G_k$ explicitly. Oct 30 '19 at 10:38