This is one of my homework questions.
Q. Let $a$ be a positive real number. Define the sequence $\{x_n\}_{n=1}^{\infty}$ by $x_1=a$ and for $n\ge 1$, $x_{n+1} = x_n(x_n+\frac 1n)$
a. Suppose that $a$ is such that the sequence $\{x_n\}_{n=1}^{\infty}$ is monotone increasing and bounded. Find the value of $\lim_{n \to \infty}x_n$ (You do not have to show that such an $a$ exists (it does)).
I can solve this, so it is not a problem.
b. Show that there exists $a$ such that the sequence $\{x_n\}_{n=1}^{\infty}$ is not monotone.
Attempt: (I am not sure whether it is monotone increasing or decreasing, but I assume that I have to show it is not monotone increasing.) Then, I have to show that there exists some $n$ such that $x_n>x_{n+1}$.
Then, $x_n >x_n(x_n+\frac 1n)\rightarrow 1>x_n+\frac 1n \rightarrow x_n < 1-\frac 1n$.
So, when $n=1, x_1=a<0.$ But, the question says $a\in Z^+$. Does this mean $a$ does not exist?
c. Show that there exists $a$ such that the sequence $\{x_n\}_{n=1}^{\infty}$ is unbounded.
I have to find $M\in N$ such that $|x_n|>B$ for $n\ge M$ and $B\in R$.
Let's suppose $|x_n|$ is bounded. Then, there exists $M \ge B$ such that for $n\ge M$, $B-\varepsilon <|x_n|<B$ for $\varepsilon \in (0,1)$.
Then, I have to show $|x_{n+1}| > |\cdot|>B$. I hope to see it is a condtradiction, but don't know how to proceed from here.
Could you give me any hint?
Thank you in advance.
Best Answer
No, the question as posted only says that $\,a \in \color{red}{\mathbb{R}^+}\,$. Hints...
c. Quite obviously $\,x_{n+1} \gt x_n^2\,$, then by telescoping $\,x_{n+1} \gt x_n^{2^1} \gt x_{n-1}^{2^2}\gt \ldots \gt x_1^{2^n}=a^{2^n}\,$. It's enough to find one $\,a\,$ such that $\,a^{2^n}\,$ diverges to $\,+\infty\,$.
b. Try writing the first few terms for some particular $\,a$'s. Given c. above, it may be tempting to look at values $\,a \lt 1\,$.