Convergence of a recursive sequence $u_n$

Question

This is my question. Let the sequence $u_n$ be recursively defined by $u_{n+1} = \frac{k}{1+u_n}$ where k>0 and $u_1$>0. Check the convergence of the sequence.

So I used this method.

Since $u_1$>0 and k>0 for all n $\epsilon$ $Z_+$,$u_n$>0

$$\lim_{n\to \infty} u_{n+1} = \frac{k}{1 + \lim_{n\to \infty} u_n}$$ Suppose $\lim_{n\to \infty} u_n$ exist and $\lim_{n\to \infty} u_n = t$ then $\lim_{n\to \infty} u_{n+1} = t$

$$t = \frac{k}{1+t}$$ $$t^2 + t - k = 0$$ $$t = \frac{-1\pm\sqrt{1+4k}}{2}$$ Since $u_1$>0 and k>0 for all n $\epsilon$ $Z_+$,$u_n$>0

$$\lim_{n\to \infty} u_n = \frac{\sqrt{1+4k}-1}{2}$$ Therefore the sequence is convergent

I'm not really sure this is the write way to do this. Please help me. If there is a better way please mention it.

Answer 1

Just some thoughts:

First of all, the sequence is bounded:

\begin{align*} u_{n} = \frac{k}{1+u_{n-1}} = \frac{k}{1+\frac{k}{1+u_{n-2}}} = \frac{(1+u_{n-2}) \cdot k}{1+u_{n-2} + k} < \frac{(1+u_{n-2}) \cdot k}{1+u_{n-2}} = k. \end{align*} Also, it is oscillating in the sense that $u_n < u_{n+1} \Leftrightarrow u_{n+1} > u_{n+2}$: If, for instance, $u_{n} < u_{n+1}$, then \begin{align*} u_{n+2} = \frac{k}{1+u_{n+1}} < \frac{k}{1+u_n} = u_{n+1}, \end{align*} and vice versa. Using this oscillating property, one can prove that $|u_{n+1} - u_n | < |u_{n}- u_{n-1}|$ for all $n$, and with some estimates of the size of $|u_{n+1}-u_n|$ one can then hope to prove that

\begin{align*} \lim_{n \rightarrow \infty} |u_{n+1}-u_n| = 0. \end{align*}

This will force the sequence to zoom in on some real point, which will of course be the limit.

Answer 2

You have proven that if the limit exists, it is $\sqrt{k+\frac14}-\frac12$. However, you have not proven that the limit exists.

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Suppose that $$ u_{n+1}=\frac{k}{1+u_n}\tag1 $$ Since $k\gt0$, if $u_n\gt0$, then $u_{n+1}\gt0$.

Adding $1$ to $(1)$ and multiplying by $1+u_n$ yields $$ (1+u_n)(1+u_{n+1})=k+1+u_n\tag2 $$ Furthermore, $$ \begin{align} u_{n+1}-u_n &=\frac{k}{1+u_n}-\frac{k}{1+u_{n-1}}\tag3\\ &=\frac{k(u_{n-1}-u_n)}{(1+u_n)(1+u_{n-1})}\tag4\\ &=\frac{k(u_{n-1}-u_n)}{k+1+u_{n-1}}\tag5\\ &=-(u_n-u_{n-1})\,\frac{k}{k+1+u_{n-1}}\tag6 \end{align} $$ Explanation:
$(3)$: apply $(1)$
$(4)$: subtract fractions
$(5)$: apply $(2)$
$(6)$: rearrange factors

Thus, $$ |u_{n+1}-u_n|\le|u_n-u_{n-1}|\frac{k}{k+1}\tag7 $$ Since $\frac{k}{k+1}\lt1$, we can compute the following sum, which, because of $(7)$, is bounded by a geometric series with ratio $\frac{k}{k+1}$: $$ \sum_{n=1}^\infty|u_{n+1}-u_n|\le\overbrace{\,(k+1)\,}^{\frac1{1-\frac{k}{k+1}}}\overbrace{|u_2-u_1|}^{\text{first term}}\tag8 $$ Since the left hand side of $(8)$ converges, the sequence converges to $$ \lim_{n\to\infty}u_n=u_1+\sum_{n=1}^\infty(u_{n+1}-u_n)\tag9 $$