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Give a concrete sequence of rationals which converges to an irrational number and vice versa....

My work

I could give a sequence of irrationals which converges to a rational number...

Let $r\in \mathbb Q,$ $$a_n=\frac {\sqrt 2} n+r$$ But I couldn't give a sequence of rationals which converges to an irrational.. Help me to work out.....

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21 Answers 21

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$a_n = \left( 1 +\dfrac{1}{n} \right)^n $ converges to $e$

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    $\begingroup$ oh.. super... thank you so much...@Luis $\endgroup$ – David Nov 15 '14 at 3:35
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    $\begingroup$ @David You are welcome $\endgroup$ – FormerMath Nov 15 '14 at 19:01
  • $\begingroup$ Really nice answer. +1. $\endgroup$ – Avinash N Oct 6 '18 at 10:33
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How about $a_n$ = the decimal expansion of $\sqrt{2}$ up to the $n$-th place

A formal definition could be $$a_n = \lfloor 10^n\sqrt{2} \rfloor 10^{-n} $$

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    $\begingroup$ +1, but this is a bit like cheating. $\endgroup$ – Joao Nov 15 '14 at 4:42
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    $\begingroup$ Why cheating? It is a concrete way of producing a sequence of rationals that converges to any irrational number of your choice. $\endgroup$ – Martin Argerami Nov 15 '14 at 12:43
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    $\begingroup$ ... whereas if you've just this minute constructed the real numbers with Dedekind cuts, then it's difficult to avoid offering this answer or the equivalent in base-2 :-) $\endgroup$ – Steve Jessop Nov 16 '14 at 13:21
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    $\begingroup$ @eBusiness: What's "artificial" about it? The most common and natural definition of the real numbers is to identify them with the Cauchy sequences of rational numbers. So what you are telling me is that using a Cauchy sequence of rational numbers to construct a Cauchy sequence of rational numbers is artificial. $\endgroup$ – Martin Argerami Nov 16 '14 at 17:20
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    $\begingroup$ You could formulate the same sequence as $a_n := 10^{-n} \max \big\{k \in \mathbb{Z} \,:\, k^2 \le 2 \cdot 10^{2n} \big\}$. Then perhaps @eBusiness would be happier with your answer. $\endgroup$ – 6005 Nov 16 '14 at 19:20
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Take ratios of consecutive Fibonacci numbers: $\frac11,\frac21,\frac32,\frac53,\frac85,\dots$. It is well known that this converges to the golden ratio $\frac{1+\sqrt{5}}{2}$, which is irrational.

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Consider the sequence $$0.1, 0.101, 0.101001, 0.1010010001, 0.101001000100001, 0.101001000100001000001,\dots.$$ This converges to $\alpha=0.101001000100001000001\cdots$. The decimal expansion of $\alpha$ is not ultimately periodic, so $\alpha$ is irrational.

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    $\begingroup$ Space those 1's out well enough and you get a transcendental number too. Thanks to Liouville. $\endgroup$ – Joel Nov 14 '14 at 18:07
  • $\begingroup$ @Joel I think that this is transcendental, too, but I've never seen the proof. $\endgroup$ – Akiva Weinberger Nov 16 '14 at 23:04
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    $\begingroup$ If I had let the gaps grow like $n!$, the number would be for sure transcendental. With the gaps I used, I believe that whether the number is transcendental is an open problem. $\endgroup$ – André Nicolas Nov 17 '14 at 8:08
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Any Taylor series that converges to an irrational would work. E.g.

$$\frac{1}{0!}+\frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} + \cdots = e$$

Pick any irrational number, pick a function that calculates it given a rational input, then the taylor series of that function around that input will fulfill the requirements.

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    $\begingroup$ It's not quite as general as you make it out to be. Not all Taylor series have rational coefficients. $\endgroup$ – Henning Makholm Nov 15 '14 at 13:45
  • $\begingroup$ Very good point - I guess more fully it requires the function and all of its derivatives to be rational about some point (and the input to be within the radius of convergence of that point). Although this is true for most well-known Taylor series, as I think it's what they're usually used for. $\endgroup$ – Joe K Nov 17 '14 at 17:57
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    $\begingroup$ I think there should be another term ${1\over 1!}$ at LHS $\endgroup$ – Supriyo Halder Jun 8 '18 at 17:16
  • $\begingroup$ @SupriyoHalder - yes and edit has fixed this $\endgroup$ – tom Nov 12 '18 at 3:19
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$$ 1 + \cfrac 1 {1 + \cfrac 1 {1 + \cfrac 1 {1 + \cfrac 1 {1 + \cfrac 1 \ddots}}}} $$

The simple continued fraction expansion of a rational number must terminate because you can't keep getting smaller and smaller positive integers, thus, for example: \begin{align} \frac{67}{30} & = 2 + \frac 7 {30} & & (\text{7 is smaller than 30}) \\[10pt] & = 2 + \frac 1 {\left(\frac{30} 7 \right)} = 2 + \cfrac 1 {4 + \cfrac 2 7} & & (\text{2 is smaller than 7}) \\[10pt] & = 2 + \cfrac 1 {4 + \cfrac 1 {\left(\frac 7 2 \right)}} = 2 + \cfrac 1 {4 + \cfrac 1 {3 + \cfrac{1}{2}}} & & (\text{1 is smaller than 2}) \end{align}

Hence the thing that an infinite simple continued fraction converges to must be irrational.

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You can get this result using Fourier series:

$$ \pi = 4\sum\limits_{n=1}^\infty \frac{(-1)^{n+1}}{2n - 1} $$

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    $\begingroup$ This seems like overkill for "my first convergent sequence". $\endgroup$ – Ben Millwood Nov 15 '14 at 15:12
  • $\begingroup$ Or Taylor? $ \ $ $\endgroup$ – BCLC Jul 17 '18 at 3:20
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Let $a_1=1$ and recursively $a_{n+1}=\frac{a_n}2+\frac{1}{a_n}$ and show that this converges to $\sqrt 2$.

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Set $a_0=1$ and $$ a_{n+1}=\frac{3a_n+4}{2a_n+3} $$ and prove the sequence is increasing and bounded; then prove that it converges to $\sqrt{2}$.

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Take any positive number $\alpha \notin \mathbb Q$. Then the rational sequence $a_n = \dfrac{\lfloor n\alpha \rfloor}{n}$ converges to $\alpha$.

(This is also true for $\alpha \in \mathbb Q$, but that case is not relevant to your question.)

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One way to go about this is to use newton's method combined with polynomials.

For instance, we know that the roots of $$f(x) = x^2 - 2$$ are $\pm \sqrt{2}$. We seed our sequence with $2$ as a first guess. The next guess comes from newton's method, this is where we compute the linearization of $f(x)$ at $2$ and find where the linearization is zero. This becomes our new guess: $$L(x) = f'(2)(x-2)+f(2) = 4(x-2) + 2$$ and we see that $L(x) = 0$ when $x = 1.5$. Thus the next term in the sequence is $1.5$.

Given a point $x_n$ we can find the next point $x_{n+1}$ via: $$L(x_{n+1}) = 0 = f'(x_n)(x_{n+1}-x_n)+f(x_n).$$ Solving for $x_{n+1}$, we have $$x_{n+1} = \frac{-x_n^2+2}{2x_n} + x_n = \frac{x_n^2+2}{2x_n} = \frac{x_n}{2} + \frac{1}{x_n}.$$

It isn't hard to prove convergence of the series, given that you choose good initial conditions. (Say $x_0 \ge 2$) Notice that this is the same sequence proposed by @HagenvonEitzen.

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    $\begingroup$ Notice if $x_0$ is rational, then so is every member of the sequence. $\endgroup$ – Joel Nov 14 '14 at 17:58
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    $\begingroup$ You can construct a sequence of rational numbers that converge to $\sqrt{a}$ for any $a\in \mathbb{N}$ using this approach. $\endgroup$ – Joel Nov 14 '14 at 18:00
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Take any irrational number, and look at its decimal expansion, like $$\sqrt{2}= 1.41421356237310\ldots$$

A sequence that converges to it is

$$1, 1.4, 1.41, 1.414, 1.4142, 1.41421, \ldots$$

Basically, just add a digit each time. This is clearly a sequence of rationals because every decimal number with a finite decimal expansion is rational, and it clearly converges to the irrational number in question (if it didn't, then decimal expansions would not be well-defined, but this can also be proved directly using geometric series).

This seems like a trivial example, but understanding it is really fundamental to understanding how decimal expansions work, and what infinite decimal expansions really are.

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There have been a lot of elegant and precise examples in response to this question, but I want to show that actually, you don't need elegance or precision: there are loads of possible answers to this kind of question.

Start with your candidate number, say $\sqrt 2$. Choose some range about it, say $\sqrt{2} \pm 1$. Now, that's a reasonable chunk of the number line, so you must be able to find a rational number in it, because rational numbers are packed in everywhere. $3/2$ will do, that's the first element of your sequence. Now choose a smaller range, say $\sqrt{2} \pm 1/2$. It turns out $3/2$ fits in this one too, so let's pick that one again. Now choose $7638$. That's nowhere near where you want to be, but it actually doesn't matter – if you only make finitely many "mistakes" like that, it doesn't make the slightest bit of difference to where you converge.

Okay, now let's pick some rational in the range $1$ to $1.5$, because that's a smaller range than before that still contains $\sqrt{2}$. Let's pick $1.23523$ (notice this is actually further away than $3/2$ was – this doesn't matter). Narrow the range to $1.25$ to $1.45$ (you can check this range is still ok by squaring both ends and checking the lower one falls below $2$ and the upper one above it). Pick another rational. Narrow again, and pick again. Keep going forever.

As long as the width of the ranges you allow eventually approach zero, the rational sequence you pick will eventually converge to $\sqrt 2$.


Here's another approach: pick a sequence that converges to $0$. Add $\sqrt{2}$ to every element. Now your sequence converges to $\sqrt{2}$, but it might not have rational elements. However, you can fix it: as before, rationals are packed in everywhere, so you ought to be able to move each element of your sequence just a tiny amount – say, less than its distance to $\sqrt{2}$, so it ends up less than twice as far away – and hit a rational number nearby. Now all your sequence members are rational, and you're done.

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  • $\begingroup$ This is a nice and intuitive procedure which most students should be able to follow. However, the digression taken by "choose $7638$" I think will just lead to confusion (among new students). Why do it, when you don't need to? $\endgroup$ – Winther Nov 16 '14 at 17:35
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    $\begingroup$ @Winther: I guess you might have a point. I'm just trying to illustrate that most of the choices don't matter at all, and it doesn't have to be "neat" or smooth or anything. $\endgroup$ – Ben Millwood Nov 16 '14 at 18:18
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question 1: $3, 3.1, 3.14, 3.141, ...$

question 2: $3.14159..., 0.14159..., 0.04159...., 0.00159, ...$

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$4*(\frac11-\frac13+\frac15-\frac17\dots)$ is known to converge to $\pi$ (Ramanujan).

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    $\begingroup$ It appears it was known long before Ramanujan: en.wikipedia.org/wiki/Leibniz_formula_for_%CF%80 One of the references from the Wiki claims it appeared in a 1670 publication by James Gregory and rediscovered by Leibniz four years later in 1674. (And presumably re-rediscovered by Ramanujan a few hundred years later!) $\endgroup$ – Hugh Denoncourt Nov 16 '14 at 8:27
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Let $x_n = {1 \over n}\sqrt{2}$. $x_n \rightarrow 0.$ $x_n$ is irrational for each $n \in \Bbb N$ but $0$ is rational.

Consider the decimal expansion of $\sqrt 2 = 1.414213562$; let $q_n$ but that decimal expansion truncated to $n$ decimal places.

Alternatively, you (may) know that $\sin(x) = \sum_{r=0}^\infty {(-1)^r x^{2r+1} \over (2r+1)!}$; let $s_n = \sum_{r=0}^n {(-1)^r ({1 \over 3}\pi)^{2r+1} \over (2r+1)!}$. $s_n \rightarrow \sin({1 \over 3}\pi) = {1 \over 2}\sqrt{2}$.

Hopefully these are slightly different to some of the previous answers! :)

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  • $\begingroup$ $s_n$ is a sequence of irrational numbers that converge to a irrational number. However, if you take $\pi/3 \to \pi/2$ instead then this would work, but you also need to know that the sum of $\text{rational}\times \pi^{2r+1}$ yields a irrational number (which requires knowledge that $\pi$ is trancendental). $\endgroup$ – Winther Nov 16 '14 at 17:39
  • $\begingroup$ Oh yeah, oops. Haha! $\endgroup$ – Sam T Nov 16 '14 at 23:03
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Let $x$ be the irrational number you are trying to obtain as the limit of a sequence of rationals. Let's assume that you can calculate the decimal expression of $x$ to the $n$th term.

Then, you can define a sequence of rationals where the $n$th term is the number $x$ with $n$ decimal digits. A number with a finite number of decimal digits is a rational number (you can multiply and divide by ten to the $n$, being $n$ the number of decimals).

It's obvious that this sequence converges to your irrational number, and the definition can easily be formalized.

With this method you can obtain ANY irrational number as long as you can calculate the $n$th decimal term with an algorithm.

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Quite a few previous answers have give sequences of rationals converging to an irrational. Here's a sequence of irrational numbers converging to a positive rational $r$:

Let $a_0=0$ and $s_i=\Sigma_0^i a_i$ and $a_{i+1}= (r-s_i)/\pi$.

(A flaw in this method: some values among the $a_i$ might by chance be rational; the sum of two irrationals need not be irrational. To address this flaw, I need to add a rational fraction of an irrational to $a_i$, rather than adding an irrational fraction of an irrational to $a_i$. This can be done by a technique similar to that in the next part of this answer.)

A sequence of rationals converging to a positive irrational $x$ can be created in a related way:

Let $a_0=0$ and $a_{i+1} = a_i + 1/k$, with $k$ the largest integer such that $a_{i+1} < x$.

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Consider $x_n=\frac{\sqrt{a}\left[(\sqrt{a}+1)^{2n}+(\sqrt{a}-1)^{2n}\right]}{(\sqrt{a}+1)^{2n}-(\sqrt{a}-1)^{2n}}$.

It is rather straightforward to see that this sequence converges to $\sqrt{a}$. The 'surprise' or key point is that actually $x_n$ is rational when $a$ is rational. This follows immediately upon using the binomial expansion.

As an example consider $a=2$. One then finds $x_n=(\frac{3}{2},\frac{17}{12},\frac{99}{70},\ldots)=(1.5, 1.417\ldots,1.414\ldots,\ldots)$, already pretty good approximations of $\sqrt{2}=1.4142\ldots$

Note: This example is essentially a reworking of the Fibonnaci to golden ratio example.

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Let $c$ be an irrational number. Now we get a rational between $c-1/n$ and $c+1/n$ for all natural numbers $n$. Then, we consider the sequence $x_n$, of the rational numbers between $c-1/n$ and $c+1/n$. Then, by squeeze theorem, we get $\lim (x_n)=c$.

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  • $\begingroup$ Hi, I edited a bit your answer (please, use Latex for mathematical symbols), but I do think it is better you rephrase the it, because it does not look clear, at least to me. Good luck! $\endgroup$ – user559615 Nov 7 '18 at 10:54
  • $\begingroup$ The question is about concrete sequences, no? $\endgroup$ – epimorphic Nov 7 '18 at 14:17
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$pi=3.14159265358979323846264338327950288419716939937510 58209749445923078164062862089986280348253421170679 82148086513282306647093844609550582231725359408128 48111745028410270193852110555964462294895493038196 44288109756659334461284756482337867831652712019091 45648566923460348610454326648213393607260249141273 72458700660631558817488152092096282925409171536436 78925903600113305305488204665213841469519415116094 33057270365759591953092186117381932611793105118548 07446237996274956735188575272489122793818301194912 98336733624406566430860213949463952247371907021798 60943702770539217176293176752384674818467669405132 00056812714526356082778577134275778960917363717872 14684409012249534301465495853710507922796892589235 42019956112129021960864034418159813629774771309960 51870721134999999837297804995105973173281609631859 50244594553469083026425223082533446850352619311881 71010003137838752886587533208381420617177669147303 59825349042875546873115956286388235378759375195778 18577805321712268066130019278766111959092164201989...$

Let's construct a sequence $3,3.1,3.14,3.141,...$ which is converges to $pi$. And this sequence digits are rational numbers and point of convergence is irrational (transcendental) number.

For another idea, just consider the sequence of partial sums of an infinite series $e^1$. It also works.

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