# Prove that the sequence is not monotone and not bounded.

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?

• How did you get $x_1 = a < 0$ ? The inequality was required to hold for some $n$, not all $n$ including $1$ . Also, "not monotone" could mean alternating increasing and decreasing. – mr_e_man Apr 28 '18 at 3:22

But, the question says $a\in Z^+$.

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 $\,a \lt 1\,$.

• The only solution to $x_2 = x_1$ is $x_1 = 0$, which contradicts $a > 0$ . – mr_e_man Apr 28 '18 at 3:42
• @mr_e_man Right, thanks. Just edited that part out, I had it off-by-one in my draft calculations. – dxiv Apr 28 '18 at 3:45
• I agree that it should be $a\in R^+$Regarding your hint b, it shows the case where $x_n$ is monotone decreasing. Am I right? – Sihyun Kim Apr 28 '18 at 3:56
• @SihyunKim -- Try $a = x_1 = 0.1$ . Then $x_2 = 0.11$ , and $x_3 = 0.0671$ . That's an increase and a decrease, so the sequence is not monotone. – mr_e_man Apr 28 '18 at 4:01
• @SihyunKim Point b. is "Show that there exists a such that ..." so, yes, once you showed that for example $\,a=0.1\,$ produces a sequence which is not monotonic, that answers the question. More generally, one counterexample is enough in order to prove a proposition false. – dxiv Apr 28 '18 at 4:10

For any $y>0$ the quadratic formula shows that $a^2+a=y$ has a positive solution for $a$. So we can choose $y$ to be any positive real and there will exist $a=a_1>0$ such that $a_2=a_1^2+a_1=y.$

Obviously $a_2>a_1$ if $a_1>0.$ So we can violate monotonicity of the sequence $(a_n)_n$ if we can find $y$ such that $y>y^2+y/2.$

If $y>0$ then $y>y^2+y/2\iff 1>y+1/2\iff 1/2>y.$ So let $a=\frac {-1+\sqrt{1+4y}}{2}$ where $0<y<1/2.$ That is, $0<a<\frac {\sqrt 3\;-1}{2}.$