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.