Sequence problem involving inequality Consider a sequence defined by $\{a_k\}_{k\geq 0}$ , $a_{k+1}=2^k-3a_k$; Find all $a_0$ such that $a_0<a_1<a_2<a_3<\cdots$.
I tried to create some bound on the terms but it doesn't satisfy me, any suggestion would be highly helpful, thanks!
 A: Do you know how to solve linear recurrence relations? We're basically looking at an inhomogeneous one of those. It's a lot like a linear differential equation with constant coefficients.
Considering $a_{k+1} + 3a_k = 2^k$ in this way, you first solve the corresponding homogeneous recurrence:
$a_{k+1} + 3a_k=0$
This is solved by the sequence $a_k=A(-3)^k$ for any real $A$. To account for the inhomogenous term, we consider sequences of the form $a_k=A(-3)^k + B(2)^k$. Plugging this into the original recurrence leads to $B=\frac{1}{5}$. Putting in an initial condition, we then find that $A=a_0-\frac{1}{5}$. Thus the formula for our sequence is:
$a_k=\left(a_0-\frac{1}{5}\right)(-3)^k + \frac{1}{5}(2)^k$.
No matter how small a non-zero coefficient we have in front of that oscillating term, it will eventually drown out the growth term. Therefore, the sequence will only keep increasing if $a_0-\frac{1}{5}=0$, i.e., if $a_0=\frac{1}{5}$.
A: Some playing around with Maple gives the following conjecture: the inequality $a_k<a_{k+1}$ gives an upper bound for $a_0$ when $k$ is even, a lower bound when $k$ is odd, and these bounds converge to a unique solution
$$a_0=0.012101210121\cdots$$
in base $3$, that is, $a_0=\frac{16}{80}=\frac{1}{5}$ only.  Would love to see if anyone can prove (or disprove) this.
Update.  @GTonyJacobs has done so.
A: Here is a different approach to computing $a_k$.
Let $b_k=a_k(-3)^{-k}$, then we have
$$
a_{k+1}=2^k-3a_k\implies(-3)^{k+1}b_{k+1}=2^k-3\cdot(-3)^kb_k
$$
Dividing by $(-3)^{k+1}$, we get
$$
b_{k+1}=-\frac13\left(-\frac23\right)^k+b_k
$$
We can compute $b_k$ using the formula for the sum of a geometric series:
$$
b_k=b_0-\frac15\left(1-\left(-\frac23\right)^k\right)
$$
Back out the change of variables to get $a_k$
$$
\begin{align}
a_k
&=(-3)^ka_0-\frac15\left((-3)^k-2^k\right)\\
&=\frac152^k+\left(a_0-\frac15\right)(-3)^k
\end{align}
$$
As has been noted, this means that the only initial $a_0$ that gives a monotonically increasing sequence $a_k$ is $a_0=\frac15$. For any other value of $a_0$, the factor of $(-3)^k$ will cause oscillation.
A: Note that:
$a_{k+1}-a_k=2^k-4a_k=4(2^{k-2}-a_k)$. 
We want $a_0$ such that $a_{k+1}-a_k>0$ for all $k$.
Now can you finish the problem?
