I'm solving some previous calculus exams, and this question came up:
Let $(a_n)$ be a sequence that defined as $a_1=1/2$, $a_{n+1}=a_n^2+a_n^3$.
Let $(b_n)$ be a sequence that defined as $b_1=-1$, $b_{n+1}=b_n^2+b_n^3$.
Let $(c_n)$ be a sequence that defined as $c_n = \left\{\begin{matrix}
a_n & n \; odd\\
b_n & n \; even
\end{matrix}\right.$Prove that $(a_n)$, $(b_n)$ are converging, is $(c_n)$ also converging?
My Solution:
We'll prove that $(a_n)$ is a converging sequence. We'll show by induction that $(a_n)$ is monotonically decreasing:
Base case: $n=1$: $a_1 = 1/2 > 3/8 = a_1^2+a_1^3$.
Induction step: For some n: $a_n > a_{n+1}= a_n^2+a_n^3$. (I'm having trouble continuing this step)
We'll check if the sequence is bounded, since the sequence is monotonically decreasing
then the upper bound is $a_1=1/2$. Noticing that $a_{n+1}=a_n^2+a_n^3$ is always a positive number, therefore: $a_n^2+a_n^3 > 0$ (Not sure how correct this is.).Since we know that $a_n$ is bounded and monotonically decreasing then we know that $lim_{n->\infty} a_n=inf\{a_n | n \in \mathbb{N}\}$, which means $lim_{n->\infty} a_n=0$.
Notice that we can shift (Not sure if this is the correct term) $(b_n)$ by $b_1=b_1 * -1/2=a_1$ therefore its limit is the same as $(a_n)$: $lim_{n->\infty} a_n=lim_{n->\infty} b_n=0$.
Therefore $(a_n)$ and $(b_n)$ are converging sequences.
We need to determine now whether $(c_n)$ is a converging sequence.
Finding the limit of $(c_n)$: $lim_{n->\infty} c_{2n-1} = lim_{n->\infty} a_n=0$ and $lim_{n->\infty} c_{2n} = lim_{n->\infty} b_n=0$ therefore $lim_{n->\infty} c_n=0$ is converging.
There are some parts that I pointed out in the proof where I had troubles getting to. Also is my proof okay? is there an easier way to approach this?
Best Answer
To show that $\{a_n\}$ is a convergent sequence it is enough to show that it is bounded below and monotonically decreasing (like you did, but didn't fully complete). We can show that sequence $\{a_n\}$ is monotonically decreasing and then the limit will give you the lower bound of the sequence.
Using mathematical induction what you did is showed for $n=1$, $a_1=\frac{1}{2}>\frac{3}{8}=a_2$. To show that $a_{n}>a_{n+1}=a_n^2+a_n^3,\forall n\in \mathbb{N}$, since $0<a_n\le \frac{1}{2}$ we can use the fact that $a_n\ge2a_n^2,\forall n\in \mathbb{N}$ (you can prove this as an exercise) and $a^2_n>a_n^3$. From here it is obvious that $a_n\ge2a_n^2>a_n^2+a_n^3,\forall n\in \mathbb{N}$. This means that the sequence is monotonically decreasing. You already found its lower bound and as you have shown $\displaystyle\lim_{n\to\infty}{a_n}=0$.
Showing the convergence of sequence $\{b_n\}$ is almost trivial. From the fact that $b_1=1$ it is obvious that $b_2=1+(-1)=0$, hence $b_n=0,\forall n\ge2$. This means $\displaystyle\lim_{n\to\infty}{b_n}=0$.
The convergence of $\{c_n\}$ you've done correctly.