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$$ F_k=\begin{cases} 4G_{k-1},\ k>1\\ 0,\ k=1 \end{cases} \\ G_k=\begin{cases} 4^{k-1}-2G_{k-1},\ k>1\\ 1,\ k=1 \end{cases} $$ 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.

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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.

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  • $\begingroup$ Wow, it was a simple trick! Now I see. Thank you! $\endgroup$
    – Bonrey
    Oct 30 '19 at 10:17
  • $\begingroup$ @Bonrey Yes now it remains to show that the limit exists. See also my edit. $\endgroup$
    – Robert Z
    Oct 30 '19 at 10:22
  • $\begingroup$ But how can I prove that L exists if $\frac{G_k}{4^k}$ is not monotonous and bounded at the same time? $\endgroup$
    – Bonrey
    Oct 30 '19 at 10:34
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    $\begingroup$ $a_k=\frac{G_k}{4^k}$ is bounded but not monotone. Consider the subsequence $a_{2k}$ and $a_{2k+1}$. Or see my P.S. and find $G_k$ explicitly. $\endgroup$
    – Robert Z
    Oct 30 '19 at 10:38

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