It is given that $u_{n+1} =1+\dfrac{1}{u_n}$ and $u_1 =1$. Find the limit of $u_n$ as $n\to\infty$.
The limit is $\frac{\sqrt{5}+1}{2}$ from a calculator. Is there an algebraic way to determine this? You can also determine that sequence is bounded but not monotonic.
If you want to be rigorous you should proove that the limit exists.
You can do that showing that the odd terms form a decreasing sequence while the even terms form an increasing sequence (induction). Then show that those sequences are bounded below/above.
Writing $u_n$ as $\frac{p_n}{q_n}$ for two sequences $(p_n)$, $(q_n)$ to be determined. We have:
$$u_{n+1} = 1 + \frac{1}{u_n} \iff \frac{p_{n+1}}{q_{n+1}} = 1 + \frac{q_n}{p_n} = \frac{p_n+q_n}{p_n}$$ Normalize $(p_n)$ and $(q_n)$ appropriately, we can turn the non-linear recurrence equation into a linear one:
$$\begin{pmatrix}p_{n+1}\\q_{n+1}\end{pmatrix} = \begin{pmatrix}1&1\\1&0\end{pmatrix}\begin{pmatrix}p_n\\q_n\end{pmatrix}$$ The charateristic polynomial of the matrix $\begin{pmatrix}1&1\\1&0\end{pmatrix}$ is equal to:
$$\lambda (\lambda - 1 ) - 1^2 = \lambda^2 - \lambda - 1 = (\lambda - \varphi)(\lambda + \varphi^{-1})$$ where $\varphi = \frac{1+\sqrt{5}}{2}$ is the golden ratio. The matrix has eigenvalues $\varphi$ and $-\varphi^{-1}$ with corresponding eigenvectors:
$$\begin{pmatrix}1&1\\1&0\end{pmatrix}\begin{pmatrix}\varphi\\1\end{pmatrix} = \varphi \begin{pmatrix}\varphi\\1\end{pmatrix} \quad\text{ and }\quad \begin{pmatrix}1&1\\1&0\end{pmatrix}\begin{pmatrix}-\varphi^{-1}\\1\end{pmatrix} = -\varphi^{-1}\begin{pmatrix}-\varphi^{-1}\\1\end{pmatrix} $$ Since $u_1 = 1$, we will choose $p_1 = q_1 = 1$. Expressing $\begin{pmatrix}1\\1\end{pmatrix}$ in terms of the eigenvectors:
$$\begin{pmatrix}1\\1\end{pmatrix} = \alpha \begin{pmatrix}\varphi\\1\end{pmatrix} + \beta \begin{pmatrix}-\varphi^{-1}\\1\end{pmatrix} \implies \alpha = \frac{\varphi+1}{\varphi+2} \;\text{ and }\; \beta = \frac{1}{\varphi+2} $$ We get $$ \begin{pmatrix}p_n\\q_n\end{pmatrix} = \begin{pmatrix}1&1\\1&0\end{pmatrix}^{n-1} \begin{pmatrix}p_1\\q_1\end{pmatrix} = \alpha \varphi^{n-1} \begin{pmatrix}\varphi\\1\end{pmatrix} + \beta (-\varphi^{-1})^{n-1}\begin{pmatrix}-\varphi^{-1}\\1\end{pmatrix}$$ This allow us to derive a closed form expression for $u_n$: $$u_n = \frac{p_n}{q_n} = \frac{\alpha \varphi^n + \beta (-\varphi^{-1})^n}{\alpha \varphi^{n-1} + \beta (-\varphi^{-1})^{n-1}} = \varphi \frac{\alpha + \beta (-\varphi^{-2})^n}{\alpha + \beta (-\varphi^{-2})^{n-1}}$$ Since $| -\varphi^{-2} | < 1$, this implies $\lim_{n\to\infty} u_n = \varphi \frac{\alpha + \beta*0}{\alpha + \beta*0} = \varphi$.
I just found out how to do it. Assume that the limit is equal to a. Therefore, the limit of u_n+1 and u_n are both equal to a. Using the recursive definition, place a back in to get. a=1+1/a which is equivalent to a^2-a-1=0. Solving for a gives two answers. But, since u_n is bounded between 1 and 2; (1+sqrt(5))/2 is the accepted limit.
If you are only interested in finding the limit, then here is an algebraic way to find it. Assume
$$\lim_{n\to \infty} u_n = a \implies \lim_{n\to \infty} u_{n+1} = \lim _{n\to \infty }(1+\frac{1}{u_n}) \implies a = 1+\frac{1}{a}\implies a^2-a-1=0 $$
$$ \implies a = \dots\,. $$
Note that, the terms of the sequence are positive.
