If $x_0 = 5$ and $x_{n+1} = x_n + \frac {1}{x_n},$ show that


This problem is taken from the list submitted for the $1975$ Canadian Mathematics Olympiad (but not used on the actual exam).

SOURCE : CRUX(Page Number 3 ; Question Number 162)

I tried writing out the first few terms :

$x_1 = 5+ \frac{1}{5} $

$x_2 = \big(5+\frac{1}{5}\big) + \big(5+\frac{1}{5}\big)^{-1} = \frac{x_0^2 + 1}{x_0} + \frac{x_0}{x_0^2 + 1} = \frac{(x_0^2 + 1)^2+x_0^2}{x_0(x_0^2+1)}$

$x_3 = \frac{(x_0^2 + 1)^2+x_0^2}{x_0(x_0^2+1)} + \frac{x_0(x_0^2+1)}{(x_0^2 + 1)^2+x_0^2} = Messy$

I tried a lot but could not find any general formula for the $n$th term. Does there even exist any?

Also it is clear that $\big(x_n + \frac{1}{x_n}\big)$ is an increasing function. So I think sequence diverges, but how can the $1000th$ term be calculated or aprroximated?

Any help would be gratefully acknowledged :).

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    Do you mean $x_{n}+\frac{1}{x_{n}}$ or $x_{n}+\frac{1}{n}$? – S.C.B. Jan 28 '17 at 13:26
  • @S.C.B. OH sorry...just a typo .. edit on its way... –  Jan 28 '17 at 13:30
  • This sounds like something you can probably do using an approximation by the continuous version of this problem, i.e. $f'(x)=\frac{1}{f(x)}$. A first approximation would be $\sqrt{2\cdot 1000 + 5} \sim 44.78$, so this may still need a little bit of tweaking. – Daniel Jan 28 '17 at 13:39
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    @Nirbhay, I love your $=Messy$ evaluation. Nice touch. – CodeLabMaster Jan 28 '17 at 18:52

2 Answers2



For all $n \ge 1$, then let us prove $$\sqrt{2n+25+\frac{1+\ln (n-1)}{2}} > x_{n} > \sqrt{2n+25}$$ PROOF

This holds for $n=1$. This can be checked numerically.

Assume this holds for $n=m$. Then, we can prove the left hand inequality through induction as $$x_{m+1}^2=x_{m}^2+2+\frac{1}{x_{m}^2}>2n+27 \implies x_{m+1} > \sqrt{2n+27}$$ Now from $x_{m+1}^2-x_{m}^2=2+\frac{1}{x_{m}^2}$, we now have $$x_{m+1}^2-x_{0}^2=\sum_{n=0}^{m-1} \frac{1}{x_{n}^2}<\sum_{n=0}^{m-1}\frac{1}{2n+25}<\sum_{n=1}^{m}\frac{1}{2n}\le \frac{1+\ln m}{2} \tag{1}$$ Thus, our claim is proved. The desired result follows from our Claim.

$(1)$: See here

NOTE: My original answer was wrong, which is the reason for this rather large edit.

José Carlos Santos
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    Amazing answer. What was the intuition behind the "claim" ?? –  Jan 28 '17 at 13:48
  • @Nirbhay I have often been thinking about radicals, so $x_{n}+\frac{1}{x_n}$ felt like it could be expressed using radicals. – S.C.B. Jan 28 '17 at 13:50
  • I would also like to know how one comes up with this kind of bound for a recursive sequence. – user1892304 Jan 28 '17 at 13:50
  • @user1892304 Intuition :) I can't explain it completely myself. – S.C.B. Jan 28 '17 at 13:52
  • @S.C.B. You did the question as if you knew its official solution. That's crazy !!! –  Jan 28 '17 at 13:55
  • @Nirbhay Not quite as crazy as you might think. I always think of math problems when I'm idle, and I always wanted to find the closed form the recurrence relation you posted. All I could succeed in is bounding it, which is how I knew the answer. Could you acept my answer? – S.C.B. Jan 28 '17 at 13:56
  • Friend, have you participated in the IMO or something? You deserve the gold! – N.S.JOHN Jan 29 '17 at 14:08
  • @N.S.JOHN I could have participaed. I watched the FKMO 2016, which picks Korean representatives for FKMO. Unfortunately, that year's FKMO was way~ too easy (so you had to get 100% or 83% to actually participate in the IMO) and I had been way to busy studying chemistry to properly prepare (so I slept through 2 thirds of it, getting only about a 70%). Hey, I saw you today. I answered your problem regarding $3x^2+x=4y^2+y$ didn't I? – S.C.B. Jan 29 '17 at 14:12
  • @yes you answered it. Seems like you have the same problems like me, I have to learn 9 other subjects in addition to math. Which Olympiad books do you practice? – N.S.JOHN Jan 29 '17 at 14:23
  • @N.S.JOHN I'm finished with Olympiads for now. Math alone was hard enough. You have to pass $3$ tests, passing through the span of a year! I also studied chemistry and physics, and got a gold medal (I could have gotten $1$st place but for $1$ measely point) and a grand award respectively. I also studied for biology, but didn't get to actually watch the exam. I studied with Korean books, but I also studied all of "Modern Chemistry" by Oxtoby and "Modern Physics" by Halliday. – S.C.B. Jan 29 '17 at 14:33
  • So you are still at school? How old are you, about to enter college? And it's awesome your cleared ChO and PhO. It's my aim too! Anyway I'm in 10 th grade – N.S.JOHN Jan 29 '17 at 14:37
  • @N.S.JOHN I am the same age as you. Well, in my language I am 17 but in yours I am 15 and a half. – S.C.B. Jan 29 '17 at 14:47
  • Let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/52684/discussion-between-s-c-b-and-n-s-john). – S.C.B. Jan 29 '17 at 14:50

$x_n^2-x_{n-1}^2=2+\frac{1}{x_{n-1}^2}$ for $n\geq1$.

Thus, $$x_{1000}^2=2\cdot1000+25+\frac{1}{x_0^2}+\frac{1}{x_1^2}+...+\frac{1}{x_{999}^2}>2025$$ and $$x_{1000}^2=2\cdot1000+25+\frac{1}{x_0^2}+\frac{1}{x_1^2}+...+\frac{1}{x_{999}^2}<2025+\frac{100}{x_0^2}+\frac{900}{x_{100}^2}<$$ $$<2025+4+\frac{900}{225}=2033<45.1^2$$

Because $x_{100}^2>225$.

Michael Rozenberg
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