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Are there any nice examples of infinite sequences of irrational numbers converging to rational numbers?

One idea I had was the sequence: $ 0.1001000010000001\cdots,0.1101000110000001\cdots,\cdots,0.1111000110000001\cdots,$ etc.

Where the first term in the sequence has ones in the place $i^2$ positions to the right of the decimal point. $(i=1,2,3,\dots)$ For the second term, we keep all the ones from the first term and add one in place of the first zero after the decimal point. We then add ones $i^3$ places after the decimal point. (The logical sum 1+1=1, i.e a number $i^2=j^3$ spaces after the decimal place has the value 1).

It is clear that this will converge to $1/9$, and I don't think the decimal expansion repeats at all.

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5 Answers 5

up vote 22 down vote accepted

Pick any irrational number $\alpha$ you like, then consider the sequence $\{x_n=\alpha/n\}_{n=1}^\infty$. Then each term of the sequence $x_n$ is irrational and it converges to zero as $n$ tends to infinity.

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How about the sequence $\sqrt{2}/n$, converging to zero?

Or, if you want to see some pattern with the decimal expansion, how about $\frac{\sqrt{2}}{10^n}$ giving 0.141421..., 0.0141421..., 0.00141421... ?

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Seems I overcomplicated things. Very nice. –  01000100 Jul 1 '12 at 8:35

Consider the sequence:

$$a_n=\sqrt\frac1{n^2+1}$$

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Although this question just wanted a sequence, does it make sense we search for a faster sequence converging to a rational number? I mean for example, $\sqrt 2/n^4$ is faster decreasing than $\sqrt 2/n$. –  B. S. Jul 1 '12 at 17:19
    
@Babak: You can always find a "faster" sequence. You can take the subsequence whose indices are $n_k=A(k,k)$ where $A$ is the Ackermann function; you can use all sort of non-computable and strange functions to go faster. Or you can just take $a_n=\sqrt[k]\frac1{n^k+1}$... –  Asaf Karagila Jul 1 '12 at 17:33
    
Thanks. I wish I could save this comment alone, not a whole question. :) –  B. S. Jul 1 '12 at 17:41
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@Babak: Open a notepad/gedit/nano/emacs window, copy and paste the content of the comment. Save on your dropbox/.git/svn server, viola! Alternatively, write a draft email with the content and keep it in a folder for email-notes. –  Asaf Karagila Jul 1 '12 at 17:50

$\large\sqrt[n]{n}$ is irrational when $n>1$, and $\lim\limits_{n\to\infty}\large\sqrt[n]{n}=1$.

See also the related question Existence of irrationals in arbitrary intervals.

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What do you mean by nice and better?an example is or right or wrong,there is no other possibility. Put a rational $q$ and take the sequence $q+\sqrt2/n$.

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