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Define $a_1=1$. If $n$ is an index for which $a_n$ has been defined, then define $$a_{n+1}= \dfrac{1+a_n}{2+a_n}.$$ An induction argument shows that ${\{a_n}\}$ is a sequence of positive numbers. So, if it converges the limit a must be $\ge 0$. By simple algebra and use of the induction argument ${\{a_n}\}$ is monotonically decreasing. Thus, by the Monotone Convergence Theorem, the sequence ${\{a_n}\}$ converges. By use of the fact that the limit of a subsequence (${\{a_{n+1}}\}$) of a convergent sequence (${\{a_n}\}$) is equal to the limit of the original sequence so $a$ is determined by solving the eqaution $$a= \dfrac{1+a}{2+a},$$ which results in two solutions $$a= \dfrac{-1 \pm \sqrt{5}}{2}.$$ We don't accept the negative one because we know that a $\ge 0$. But why the second solution $a= \dfrac{-1 - \sqrt{5}}{2}$ appears if it is not allowed? Isn't that the conclusion must predict solution not leaving choice or manipulation after the conclusion achieved?

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    The equation doesn't know we have $a_1=1$. What about if we had $a_1=(-1-\sqrt{5})/2$? – André Nicolas Jun 21 '16 at 07:34
  • @AndréNicolas, if $a_1=(-1-\sqrt{5})/2$ will the limit be $a=(-1-\sqrt{5})/2$? –  Jun 21 '16 at 07:36
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    If $a_1=(-1-\sqrt{5})/2$ so will be $a_2$ and so on... – N74 Jun 21 '16 at 07:43
  • @N74, O yes! Thanks a lot :) –  Jun 21 '16 at 07:46
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    The logic of the situation is one-way implication only. The argument shows that If $a$ is the limit of this sequence, then $a$ satisfies the equation $a=(1+a)/(2+a)$. At no point did you produce an argument saying that If $a=(1+a)/(2+a)$, then $a$ is the limit of this sequence. See this thread for a very similar discussion. Beware, many answers there do not really address the above root cause of the perceived problem. – Jyrki Lahtonen Jun 21 '16 at 07:47
  • It's the difference between a necessary condition and a sufficient condition. Are you familiar with these terms in a mathematical context? – MPW Jun 21 '16 at 08:05
  • @MPW, Yes, but I don't know what you mean how giving the initial value $a_1$ change which of the "sufficient" or "necessary" conditions? –  Jun 21 '16 at 08:12
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    L.G. You only have a necessary condition here. If $a$ is the limit, then ... If you had a sufficient condition, then you could work backwards, too. In yet other words. This time the conclusion leaves you with two choices for the limit. That is still a huge improvement to what you started with, when there were infinitely many possibilities for the limit (including that the limit might not exist at all). Therefore the conclusion was worth striving for. A finite selection of possibilities is trivial in comparison. – Jyrki Lahtonen Jun 21 '16 at 08:25
  • I mean that if $a$ is a limit of the sequence, then it satisfies your equation (the equation is a necessary condition); but if a number $a$ satisfies your equation, it needn't be a limit of the sequence (the equation is not a sufficient condition). – MPW Jun 21 '16 at 12:32
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    @L.G.: It is easy to verify that if we start at exactly $a_1=\text{the other one}$, the limit is indeed $(-1-\sqrt{5})/2$. But I think (have not checked) that if we start near $(-1-\sqrt{5})/2$, then we may not have convergence to $(-1-\sqrt{5})/2$ (repelling fixed point). – André Nicolas Jun 21 '16 at 13:34

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