If $a_1 = 1$, $a_{2n} = pa_{2n-1}$, and $a_{2n+1} = bp + a_{2n}$, $0 < p < 1$, $b>0$. What is $\lim \inf_na_n$ and $\lim \sup_na_n$?
I began looking for patterns: \begin{align*} a_1 &= 1\\ a_2 = p \times a_1 &= p\\ a_3 = bp + a_2 &= bp + p\\ a_4 = p \times a_3 &= p (bp + p)\\ a_5 = bp + a_4 &= bp + p (bp + p)\\ a_6 = p \times a_5 &= bp^2 + p^2 (bp + p)\\ a_7 = bp + a_6 &= bp + bp^2 + p^2 (bp + p)\\ a_8 = p \times a_7 &= bp^2 + bp^3 + p^3 (bp + p)\\ \end{align*}
Putting this in terms of evens and odds, we have:
Only odd terms: $a_{n+1} = bp + pa_n $
Only even terms: $a_{n+1} = p(bp + a_n) $
I am unsure where to progress from here to find the infimum and supremum.
Hint: I use what you have proven. Put $b_k=a_{2k-1}$. From your first formula (with only odd terms), you have $b_{k+1}=bp+pb_k$. It is now easy to find $b_k$: You find first a solution of the recurrence that is a constant sequence, namely $c=\frac{p}{1-p}b$, and then add the general solution of the recurrence $u_{k+1}=pu_k$, you get $b_k=\lambda p^k+\frac{p}{1-p}b$ for some constant $\lambda$. To find $\lambda$ note that $b_1=a_1=1$. Do the same for $c_k=a_{2k}$.