Is it possible to have a convergent sequence whose terms are all irrational but whose limit is rational?

  • 5
    $\begingroup$ Have you considered $\frac{\pi}n$ $\endgroup$ – s.harp Feb 12 '16 at 18:14
  • 2
    $\begingroup$ Yes because irrational numbers are dense in reals. $\endgroup$ – Gonenc Mogol Feb 12 '16 at 18:14
  • $\begingroup$ Take the sequence of partial sums of any of the series from here math.stackexchange.com/questions/1647409/… $\endgroup$ – Wojowu Feb 15 '16 at 16:27

$$\bigg\{a_{n} = \frac{\sqrt{2}}{n} \bigg\}$$


I will give a more interesting answer (I think OP wants something like that):




It's not hard to find such numbers that $\sqrt{b^2+4a}$ is rational.



Also, using Euler's continued fraction theorem we can have something like this:


Actually, I can do even better. Let $\phi$ be the golden ratio, then we have:


But we don't want $e$ to feel left out, so here is another one:


Another good one. Using the following:

$$2=e^{\ln 2}=e^{1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\dots}$$

We obtain an infinite product, converging to $2$:

$$\prod_{k=1}^{\infty} \frac{\sqrt[2k-1]{e}}{\sqrt[2k]{e}} =\frac{e \sqrt[3]{e} \sqrt[5]{e} \sqrt[7]{e} \cdots}{\sqrt{e} \sqrt[4]{e} \sqrt[6]{e} \sqrt[8]{e} \cdots}=2$$

  • $\begingroup$ Woah, why is the $\phi$ one true? $\endgroup$ – mysatellite Feb 15 '16 at 16:24
  • $\begingroup$ @Sky, use the identity $\phi^{-2}+\phi^{-1}=1$, then multiply it by $\phi^{-1}$ several times to get $\phi^{-3}+\phi^{-2}=\phi^{-1}$, $\phi^{-4}+\phi^{-3}=\phi^{-2}$, etc. and replace the terms in the first indentity. You can also use the closed form of geometric sum to prove this result $\endgroup$ – Yuriy S Feb 15 '16 at 18:02

If $q$ is any rational number at all and $n$ is a positive integer then $q+\frac 1 n \sqrt 2$ is irrational (it's a simple algebra exercise to prove that), and $\lim\limits_{n\to\infty}\left(q + \frac 1 n \sqrt 2\right) = q$.


Here is another example: $$a_n=\frac{1}{n\pi}$$


Let $q$ be a rational number, $\alpha$ be an irrational number, and $a_n$ be a sequence of integers where $a_n \to \infty$. Then the sequence $$b_n = q + \frac {\alpha} {a_n}$$ satisfies the property you asked.


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.