# limit of a recursive sequence, then show how it can be unbounded and not monotone

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.

Part a) $\;\ldots\;$ $x^2=x \rightarrow x=1$

$x^2 = x \implies x=1\,$ 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.

Part b) Show there exists an $a$ such that the sequence is not monotone.

Hint: $\;x_2=a(a+1) \gt a = x_1\,$. Try to find an $a$ such that $x_3 \lt x_2\,$.

Part c) Show there exists an $a$ such that the sequence is unbounded.

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.

• Following through part b) i get that $-1<a<-0.5$. Is it expected to violate $a$ is a positive real number or have i just done that wrong and that $a$ should still be greater than 0? – Bobbie May 9 '18 at 0:49
• @Bobbie Recheck your calculations. You should get that $\,x_3=x_2\,$ happens for a positive $a$, and $\,x_3 \lt x_2\,$ on one side of that value. – dxiv May 9 '18 at 0:51
• Thanks, I believe I've done part b) but i still dont really understand how you came to the hint in part c)? – Bobbie May 9 '18 at 4:43
• @Bobbie Start with $\,x_{n+1}=x_n\left(x_n+\frac{1}{n}\right) = x_n^2+\frac{x_n}{n} \gt x_n^2\,$. By the same argument $\,x_n\gt x_{n-1}^2\,$, so $\,x_{n+1} \gt x_n^2 \gt \left(x_{n-1}^2\right)^2 = x_{n-1}^{2 \,\cdot\, 2} = x_{n-1}^{2^2}\,$. Repeat, and it "telescopes" all the way down to $\,x_{n+1} \gt x_1^{2^{n}}=a^{2^{n}}\,$ (had the exponent off by $1$, corrected in the answer now). Then, if you choose an $\,a \gt 1\,$, the RHS diverges to $\,+\infty\,$, so the LHS is unbounded. – dxiv May 9 '18 at 5:19