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!

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

3 Answers 3


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,



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

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 $u_n > L$.



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 $a\ge 2$.

  • $\begingroup$ 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
    Feb 6, 2018 at 17:37
  • $\begingroup$ @idok Yes right. i edited. thanks. $\endgroup$ 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
    Feb 6, 2018 at 18:00
  • $\begingroup$ @idok If $a <2$, it will have no limit. $\endgroup$ Feb 6, 2018 at 18:22
  • $\begingroup$ Thank you all for your anwsers. $\endgroup$
    – albert
    Feb 7, 2018 at 1:56

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 $\,\Delta = a^2 - 4 \ge 0 \iff a \ge 2\,$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.