# Possible limits of a positive sequence $(u_n)_{n\geq0}$

I have a sequence defined as $$u_{n+1}=f(u_n)=(u_n^2+3)/4$$, and with $$u_0\geq0$$. I am asked to determine what are the possible limits of $$u_n$$ by studying $$f(x)$$ and the sign of $$f(x)-x$$. $$\begin{array}{c|ccccccccc} x & 0 & & 1 & & 2& & 3 & & \infty\\\hline f'(x) & & & & & +& & & &\\\hline f(x) &3/4 & & & &\nearrow& & & & \infty\\\hline s\left(f(x)-x\right)&& + & | & & - & & | & + & \end{array}$$ Suppose $$0\leq u_0\leq1$$, then $$\frac{3}{4}\leq \frac{u_0^2+3}{4}=u_1\leq1$$, which means that $$u_n$$ will always stay in $$[0, 1]$$.
The only thing I am sure of is that $$\lim_{n\to\infty}u_n\leq1$$, but I don't see how I can determine a possible limit other than $$1$$ here.

Suppose $$1\leq u_0\leq 3$$, then $$1\leq u_1\leq3$$, which means that $$u_n$$ will always stay in $$[1, 3]$$.
Since $$f(x)-x\leq0$$, and because of the sentence above, I am tempted to say that $$u_n$$ will go backwards until it reaches $$1$$, but I also don't see how I can put this mathematically, like the first case.

Suppose $$3\leq u_0\leq k$$, then $$3\leq u_1\leq\frac{k^2+3}{4}$$. Using the sign of $$f(x)-x$$, I know that $$k\leq\frac{k^2+3}{4}$$, and $$u_0\leq u_1$$. The upper bound keeps getting bigger, and the difference in two consecutive terms of the sequence also does, so there is no limit.

I would gladly accept any help for the first two parts, and a verification for the last one. Thank you very much!

Consider first $$0 \le u_0 \le 1$$. If $$u_0 = 1$$, you'll have $$u_i = 1$$ for all $$i \ge 0$$. Otherwise, as you showed, the upper bound for the values is $$1$$. Also, note that you have, with $$x = u_n \lt 1$$, that

\begin{aligned} f(x) - x & = \frac{u_n^2 + 3}{4} - u_n \\ & = \frac{u_n^2 + 3 - 4u_n}{4} \\ & = \frac{(u_n - 1)(u_n - 3)}{4} \\ & \gt 0 \end{aligned}\tag{1}\label{eq1A}

Thus, you have a strictly increasing sequence that has an upper bound. To determine it's limit point $$L$$, note that the differences between $$u_{n}$$ and $$u_{n+1}$$, as well as with $$L$$, become arbitrarily small. As such, you can determine what $$L$$ is by replacing $$u_{n}$$ and $$u_{n+1}$$ with $$L$$ and then solving, i.e.,

$$L = \frac{L^2 + 3}{4} \implies L^2 - 4L + 3 = (L - 3)(L - 1) = 0 \tag{2}\label{eq2A}$$

Thus, the limit would be $$L = 1$$.

You can do likewise for the case of $$1 \le u_0 \le 3$$, where if $$k = 3$$, then $$u_i = 3$$ for all $$i \ge 0$$, so the limit point would be $$3$$. Otherwise, for $$u_0 \lt 3$$, then \eqref{eq1A} shows in this case that $$f(x) - x \lt 0$$, so you have a strictly decreasing sequence which, once again, has a lower bound of $$1$$. Using \eqref{eq2A}, you can determine the limit is once again $$L = 1$$.

Finally, starting with $$u_0 \gt 3$$, as you've noted, it becomes an increasing sequence. Also, as you state, and \eqref{eq1A} confirms, as $$u_n$$ increases, the value of $$f(x) - x$$ also increases, so the difference keeps increasing faster, and thus the values, with no upper bound and, thus, no limit.

• Thank you very much, especially for $(2)$. I've first wrote that the difference becomes arbitrarily small, and wrote it, but then dismissed it as obvious and haven't though of considering the actual limit as a value. Thank you again! Mar 21, 2020 at 7:29