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Let $a>0$ a real number and $(u_n)$ the sequence defined by $$ u_{n+1} = a - \frac{1}{u_n}\text{ and } u_0 = a. $$ Question: Determine condition on the value of $a>0$ such that the sequence $(u_n)$ is always positive.

Attempt: I tried to establish a general formula of $u_n$ in order to set up a condition on $a$, by calculating $u_n$ with some $n$: $$ u_1 = a - \frac{1}{u_0} = a - \frac 1a = \frac{a^2-1}{a},\\ u_2 = a - \frac{1}{u_1}= a - \frac{a}{a^2-1}=\frac{a^3-2a}{a^2-1},\\ u_3 = a - \frac{1}{u_2} = a - \frac{a^2-1}{a^3-2a}=\frac{a^4-3a^2+1}{a^3-2a},\\ u_4 = a - \frac{1}{u_3} = a - \frac{a^3-2a}{a^4-3a^2+1}=\frac{a^5-4a^3+2a-1}{a^4-3a^2+1}. $$ But I wasn't successful, it seems that there is no general formula of $u_n$. So I would be appreciate for any suggestion of solution. Thank you!

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  • $\begingroup$ it is a General formula for this sequence $\endgroup$ Commented Feb 6, 2018 at 17:24

3 Answers 3

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Notice that if the sequence $u_n$ is positive, then it must be decreasing. This can be proven by induction.

In this case, the sequence $u_n$ is monotonic and bounded, so it has a limit $L$ that has to be positive.

From the formula,

$$L=a-\frac{1}{L}$$

Thus

$$a=L+\frac{1}{L} \ge 2$$

So a necessary condition is $a \ge 2$.

To prove that the condition is suffecient, Take $L>0$ such that $a=L+\frac{1}{L}$ and prove by induction that $u_n$ is decreasing, positive and $u_n > L$.

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hint

Let $f (x)=a-1/x $.

$f $ is increasing at $(0,+\infty) $.

$(u_n) $ is monotonic.

$u_1 <u_0$ thus it is strictly decreasing.

the sequences terms are $>0$ if the limit (if it exists) is $\ge 0$.

the limit $l $ satisfies $l=a-1/l .$ if $a\ge 2$ then

$l=(a\pm\sqrt {a^2-4})/2>0$.

If $a<2$ , $u_2<0$.

So, the condition is $a\ge 2$.

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  • $\begingroup$ I think you have a mistake. Why is $l=1-\frac{1}{l}$? it should be $l=a-\frac{1}{l}$, which has a solution for $a \ge 2$. See my answer. $\endgroup$
    – idok
    Commented Feb 6, 2018 at 17:37
  • $\begingroup$ @idok Yes right. i edited. thanks. $\endgroup$ Commented Feb 6, 2018 at 17:44
  • $\begingroup$ Also, the series does not tend to $-\infty$ when $a < 2$. this mistake is not crucial to the argument, but it is important not to confuse the asker. $\endgroup$
    – idok
    Commented Feb 6, 2018 at 18:00
  • $\begingroup$ @idok If $a <2$, it will have no limit. $\endgroup$ Commented Feb 6, 2018 at 18:22
  • $\begingroup$ Thank you all for your anwsers. $\endgroup$
    – albert
    Commented Feb 7, 2018 at 1:56
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Hint: $\;\require{cancel} u_{n+1} - u_n = \left(\cancel{a} - \dfrac{1}{u_n}\right) - \left(\cancel{a} - \dfrac{1}{u_{n-1}}\right)=\dfrac{u_n-u_{n-1}}{u_nu_{n-1}} \,$, so as long as the terms are positive the differences between consecutive terms have the same sign i.e. the sequence is monotonic, and is in fact decreasing since $\,u_1-u_0=\left(\cancel{a} - \dfrac{1}{a}\right) - \cancel{a} \lt 0\,$. For it to be positive it is necessary and sufficient that $\,0\,$ is a lower bound, so the sequence is positive iff it is convergent. In this case, passing the recurrence relation to the limit, its limit $\,A\,$ must satisfy $\,A = a - \dfrac{1}{A}$ $\iff A^2 - aA + 1 = 0\,$. The latter equation has real roots iff $\,\Delta = a^2 - 4 \ge 0 \iff a \ge 2\,$.

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