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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.

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    $\begingroup$ To prove that a sequence is convergent you cannot start assuming that it is convergent. $\endgroup$ – Kavi Rama Murthy Jun 6 '18 at 10:27
  • $\begingroup$ So what would be the best way to approach this question then $\endgroup$ – K.H.P.Kariyawasam Jun 6 '18 at 11:07
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    $\begingroup$ Your argument needs two parts. One part, which you did, shows if it is convergent, then the limit is what you computed. The other part needs to show that it converges. $\endgroup$ – GEdgar Jun 6 '18 at 12:41
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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.

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    $\begingroup$ I am stucked at proving $|u_{n+1} - u_n | < |u_{n}- u_{n-1}|$. Can you please elaborate your answer ? $\endgroup$ – Nimantha Jun 29 '20 at 3:56
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You have proven that if the limit exists, it is $\sqrt{k+\frac14}-\frac12$. However, you have not proven that the limit exists.


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 $$ \lim_{n\to\infty}u_n=u_1+\sum_{n=1}^\infty(u_{n+1}-u_n)\tag9 $$

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  • $\begingroup$ I understand up to $(7)$. But after that how did you get $(8)$ and prove the convergence ? $\endgroup$ – Nimantha Jul 4 '20 at 3:26
  • $\begingroup$ @Nimantha: I have expanded the explanation. Let me know if there is still anything unclear. $\endgroup$ – robjohn Jul 4 '20 at 4:25

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