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

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    $\begingroup$ Have you considered $\frac{\pi}n$ $\endgroup$ – s.harp Feb 12 '16 at 18:14
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    $\begingroup$ 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
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$$\bigg\{a_{n} = \frac{\sqrt{2}}{n} \bigg\}$$

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

$$\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+\cdots}}}}=2$$

Generally

$$\sqrt{a+b\sqrt{a+b\sqrt{a+b\sqrt{a+\cdots}}}}=\frac{1}{2}(b+\sqrt{b^2+4a})$$

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

Also:

$$\sqrt{2}^{\sqrt{2}^{\sqrt{2}^...}}=2$$

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

$$1=\cfrac{\pi^2/9}{2-\pi^2/9+\cfrac{2\pi^2/9}{12-\pi^2/9+\cfrac{12\pi^2/9}{30-\pi^2/9+\cfrac{30\pi^2/9}{56-\pi^2/9+\cdots}}}}$$


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

$$1=\frac{1}{\phi^2}+\frac{1}{\phi^3}+\frac{1}{\phi^4}+\frac{1}{\phi^5}+\cdots=\sum^{\infty}_{k=2}\frac{1}{\phi^k}$$

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

$$1=\cfrac{e}{e+\frac{1}{e}-\cfrac{1}{e+\frac{1}{e}-\cfrac{1}{e+\frac{1}{e}-\cdots}}}$$


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 $2$:

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

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  • $\begingroup$ 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
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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 $\lim\limits_{n\to\infty}\left(q + \frac 1 n \sqrt 2\right) = q$.

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

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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.

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