So the question is given a sequence $\{x_n\}_{n=1}^\infty$ given by $x_1=a$ for $a$ as a positive real number and $n\ge1$: $$x_{n+1}=x_n(x_n+\frac1n)$$ Part a) Suppose a is such that the sequence is monotone increasing and bounded. Find the value of $\lim_{n\rightarrow\infty} x_n$ For this I assumed by Monotone bounded sequence theorem the limit exists. I supposed that $$\lim_{n\rightarrow\infty}x_{n+1}=\lim_{n\rightarrow\infty}x_n$$ and the $$\lim_{n\rightarrow\infty}x_n=x$$ Then $$\lim_{n\rightarrow\infty}x_{n+1}=\lim_{n\rightarrow\infty}(x_n(x_n+\frac1n))=x^2$$ Therefore $$x^2=x \rightarrow x=1$$ Am I correct in making my first assumption and all that follows?
Part b) Show there exists an $a$ such that the sequence is not monotone.
My main thought was either an $a<0$ or an $0<a<1$ but i cant work out how to do this.
Part c) Show there exists an $a$ such that the sequence is unbounded.
For both part b and c i dont really understand where to begin or how to prove it.
$x^2 = x \implies x=1\,$ or $\;x=0\,$. In order to justify that the limit cannot be $\,x=0\,$ you need to use the positivity of $a$ and the "monotone increasing" assumptions, which leaves $\,x=1\,$ indeed.
Hint: $\;x_2=a(a+1) \gt a = x_1\,$. Try to find an $a$ such that $x_3 \lt x_2\,$.
Hint: $\;x_{n+1} \gt x_n^2 \gt x_{n-1}^{2^2} \gt \ldots \gt x_1^{2^{n}}=a^{2^{n}}\,$. Try to find an $\,a\,$ such that $\,a^{2^{n}}\,$ diverges.