Determining the limit of the sequence $a_{n+1}=-ta_n^2+(t+1)a_n$ Let $t\in(0,1]$, $a_0\in (0,\frac{1}{t}+1)$ and $a_{n+1}=-ta_n^2+(t+1)a_n$.
Does $\lim\limits_{n\to\infty}a_n$ exist for all $\,t\in(0,1]\,$ and $\,a_0\in\big(0,\frac{1}{t}+1\big)\,$? If yes, determine the limit.
I already showed that $a_n$ is bounded and I think the limit exists and is one. Now I have problems to show that $a_n$ is monotone increasing/decreasing, because it is either increasing or decreasing or nothing (for $t=0.8$ and $a_0=2.24$) it is not decreasing and not increasing.
I also tried the Banach-fixed-point theorem but I think I can't use it here.
 A: 
$\def\paren#1{\left(#1\right)}\def\N{\mathbb{N}}$Step 1: $0 < a_n < \dfrac{t + 1}{t}$ for $n \geqslant 0$.

Proof: To prove by induction, note that the case of $n = 0$ is already given. Assume the proposition holds for $n$, then by the induction hypothesis,$$
a_{n + 1} = ta_n \paren{ \frac{t + 1}{t} - a_n } > 0,
$$
and$$
a_{n + 1} = -t\paren{ a_n - \frac{t + 1}{2t} }^2 + \frac{(t + 1)^2}{4t} \leqslant \frac{(t + 1)^2}{4t} < \frac{t + 1}{t}.
$$
End of induction.

Step 2: $\lim\limits_{n → ∞} a_n = 1$.

Proof: For $n \geqslant 0$, Step 1 implies that $-t < 1 - ta_n < 1$, then $|1 - ta_n| < 1$. Therefore,$$
|a_{n + 1} - 1| = |-ta_n^2 + (t + 1)a_n - 1| = |1 - ta_n| \cdot |a_n - 1| \leqslant |a_n - 1|.
$$
This shows that $\{|a_n - 1|\}$ is a decreasing sequence, so $\lim\limits_{n → ∞} |a_n - 1| = l$ for some $l \geqslant 0$.
Suppose $l ≠ 0$, then making $n → ∞$ in$$
|a_{n + 1} - 1| = |1 - ta_n| \cdot |a_n - 1|
$$
yields $\lim\limits_{n → ∞} |1 - ta_n| = 1$. If $t = 1$, then this means that $l = 0$, a contradiction. Thus $t < 1$ and there exists $N \in \N$ such that $|1 - ta_n| > \dfrac{1}{2} (t + 1)$ for any $n > N$.
If $1 - ta_{n_0} < 0$ for some $n_0 > N$, then$$
1 - ta_{n_0} < -\dfrac{1}{2} (t + 1) < -t,
$$
a contradiction. Therefore $|1 - ta_n| = 1 - ta_n$ for $n > N$, and$$
\lim_{n → ∞} (1 - ta_n) = \lim_{n → ∞} |1 - ta_n| = 1 \implies \lim_{n → ∞} a_n = 0 \implies l = \lim_{n → ∞} |a_n - 1| = 1.
$$
Note that $\{|a_n - 1|\}$ is decreasing, thus $|a_0 - 1| \geqslant l = 1$, which contradicts $0 < a_0 < \dfrac{t + 1}{t}$.
Therefore $l = \lim\limits_{n → ∞} |a_n - 1| = 0$, then $\lim\limits_{n → ∞} a_n = 1$.
A: 
Statement: either $1\le a_{n+1}\le a_n<\frac{1}{t}$ or $0< a_{n}\le a_{n+1}\le1$.


Proof: if $0<a_n\le1$, then $1\le 1+t-ta_n<1+t$ then since $a_{n+1}=a_n(1+t-ta_n)$, we have$$
0<a_n\le a_{n+1}\le 1.
$$
If $1\le a_n<1+\frac{1}{t}$, then $0< 1+t-ta_n\le 1$ then we have$$
1\le a_{n+1}\le a_n<1+\frac{1}{t}.
$$

Proving the convergence of $a_n$ to $1$ is easy now.
A: Your expectation that the sequence always converges and its limit is $1$ is true: Either the sequence constantly equals $1$, or it it is strictly increasing or decreasing from $a_1$ onwards, depending on whether $a_1$ is smaller or larger than $1$.
The change of variables $\,z_n= t-ta_n\,$ where $\,n=0,1,2,\ldots\,$ ("$z$" like zero) implies that
$$\,z_{n+1} \,=\, (z_n+1-t)\,z_n\;\text{ and }\;-1<z_0<t\,.\tag{1}$$
If $\,z_0= 0\,$ then $\,z_n= 0\,$ for all $\,n\,$, which corresponds to the constant sequence $(a_n=1)_n\,$.
In the sequel, $\,(z_n= 0)_{n\geqslant 0}\,$ shall be excluded. We have the bounds
$$t-1\:<\: z_n \:<\: t\,\;\text{ for }\:n=1,2,\ldots$$
which are shown inductively:
Assume $\,t-1< z_n<t\,$ for some $\,n\geqslant 1$. Then $\,0< z_n+1-t<1$, which is the first factor in the recursion equation in $(1)$, hence
$$t-1\:<\: (z_n+1-t)\,z_n \,=\, z_{n+1} \:<\: t\tag{2}$$
For $\,z_1=z_0^2+(1-t)z_0\,$ these bounds result from adding $\,0< z_0^2<t^2$
and $\,-(1-t)\leqslant (1-t)z_0\leqslant t\,(1-t)$, both estimates are obtained from $(1)$.
Furthermore, it follows from $(2)$ that $(z_n)_{n\geqslant 1}$ is a strictly decreasing sequence if $z_1>0$, and it is strictly increasing
if $\,z_1<0\,$. This proves convergence.
Now going to the limit $\,z\,$ yields $\,z=z\cdot(z+1-t)\,$ from  $(1)$, thus $\,z=0\,$. If it were nonzero, then $\,z=t$, in contradiction to the strict decreasing.
In summary we get that $\big(a_n=1-\frac{z_n}t\big)_{n\geqslant 1}$ is strictly increasing and approaches $1$ from below if $\,a_1<1$.
Otherwise $(a_n)_{n\geqslant 1}$ is constant if $\,a_0=1=a_1\,$,
or strictly decreasing and approaching $1$ from above if $\,a_1>1$.
