Problem. Let \(\left(a_{n}\right)_{n=1}^{\infty}\) be a sequence s.t. \(\forall n\in\mathbb{N}\quad a_{n+1}=a_{n}\cdot(1-a_{n})\).
Prove that \(\left(a_{n}\right)_{n=1}^{\infty}\) converges if and only if \(a_{1}\in\left[0,1\right]\).
Proof. We will prove both directions separately.
\(\left(\Rightarrow\right)\) Let us assume \(\left(a_{n}\right)_{n=1}^{\infty}\) is a convergent sequence.
1. Let \(\underset{n\rightarrow\infty}{\lim}a_{n}=L\).
Since the sequence converges, the shifted sequence \(\left(a_{n+1}\right)_{n=1}^{\infty}\) must also converge to \(L\).
Taking the limit on both sides of the recurrence relation and applying sequence limits arithmetic gives \[L=L\cdot\left(1-L\right)\iff L=L-L^{2}\iff L^{2}=0\iff L=0\]
2. Note the following: \[a_{n+1}\le a_{n}\iff a_{n}\cdot(1-a_{n})\le a_{n}\iff a_{n}-a_{n}^{2}\le a_{n}\iff-a_{n}^{2}\le0\iff a_{n}^{2}\ge0\] which is true for every \(n\in\mathbb{N}\) - therefore \(\left(a_{n}\right)_{n=1}^{\infty}\) is a monotonically decreasing sequence.
3. Let us recall that every convergent sequence is bounded, and that a monotonically decreasing sequence which is bounded from below converges to the infimum of its elements.
Therefore, \(0\) is the infimum of the sequence elements and in particular \(a_{1}\ge0\).
4. If \(a_{1}=0\), then \(a_{1}\le1\). Otherwise, \(a_{1}>0\) and the infimum provides \(a_{2}\ge0\).
Substituting the recurrence relation gives \(a_{1}-a_{1}^{2}\ge0\iff a_{1}\ge a_{1}^{2}\iff1\ge a_{1}\).
5. Overall, \(a_{1}\in\left[0,1\right]\) as required.
\(\left(\Leftarrow\right)\) Assume \(a_{1}\in\left[0,1\right]\) .
By step 2 of the previous direction (this argument did not use convergence), \(\left(a_{n}\right)_{n=1}^{\infty}\) is a monotonically decreasing sequence.
Let us prove by induction that \(\forall n\in\mathbb{N}\ a_{n}\in\left[0,1\right]\).
For \(n=1\), the assumption provides \(a_{1}\in\left[0,1\right]\).
Assume for some \(n\in\mathbb{N}\) that \(a_{n}\in\left[0,1\right]\). Then, \[0\le a_{n}\le1\iff0\le a_{n}^{2}\le a_{n}\iff-a_{n}\le a_{n}^{2}-a_{n}\le0\iff a_{n}\ge a_{n}-a_{n}^{2}\ge0\iff a_{n}\ge a_{n+1}\ge0\]
The induction assumption provides \(1\ge a_{n}\) therefore \(a_{n+1}\in\left[0,1\right]\) and the induction is complete.
Let us recall that a sequence which is monotonically decreasing and is bounded from below is convergent, hence \(\left(a_{n}\right)_{n=1}^{\infty}\) is a convergent sequence. ∎