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A) a sequence $(x_n)$ of irrational numbers that converges to a rational number

Are the following answer correct? $x_n = e^1$

or

$\sqrt{2}/n$

Give a little explanation if you could

B) a sequence $(x_n)$ that is not Cauchy, but for which $|x_{n+1} - x_n|$ converges to $0$

please give an example

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  • $\begingroup$ What does your sequence $\sqrt{2}/n$ converge to (It is an example of what you want).? $\endgroup$ Nov 27, 2013 at 6:36
  • $\begingroup$ B) $a_n =|x_{n+1} - x_n|$, $|x_n - x_m| = |x_n-x_{n+1} + x_{n+1} - ... -x_m| \leq \sum_{k=n}^{m-1} a_n$ So $x_n$ is Cauchy. Hence no example. $\endgroup$
    – HK Lee
    Nov 27, 2013 at 6:41
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    $\begingroup$ @HeeKwonLee for $\epsilon>0$, can we always choose an $n$ large enough so that $\sum_{k=n}^{m-1} a_n < \epsilon$? $\endgroup$ Nov 27, 2013 at 6:47
  • $\begingroup$ You're right. I see your answer. Thank you. $\endgroup$
    – HK Lee
    Nov 27, 2013 at 6:52

2 Answers 2

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Your answer $a_n = \sqrt{2}/n$ is valid for the first part. I'm not sure what you meant to type when you put $a_n = e^1$, but as written it is not valid.

This answer to B) is one of those "classic" counterexamples that are helpful, in general, to remember. $$ x_n = \sum_{k=1}^n\frac 1k $$ Can you show that $|x_{n+1}-x_n|\to 0$? Can you show that $x_n$ is not Cauchy?

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A) The sequence $x_{n}=e^{1}$ is a constant sequence that converges to $e$, which is not rational, so this is not such an example.

The sequence $x_{n}=\frac{\sqrt{2}}{n}$ converges to $0$, because for any $\varepsilon>0$ we can pick $N$ large enough (in particular, $N=\frac{2}{\varepsilon}$ works) such that $|x_{n}-0|<\epsilon$ whenever $n>N$. Furthermore, every element in this sequence is irrational, because $\sqrt{2}$ is irrational and the product of a rational number ($\frac{1}{n}$) and an irrational number ($\sqrt{2}) is irrational. Thus this is a sequence of irrationals that converges to a rational.

B) Consider a sequence of partial sums $x_{n}=\sum_{i=1}^{n}\frac{1}{i}$. Can you tell why?

The terms $|x_{n+1}-x_{n}|=\frac{1}{n+1}$ converge to $0$ as $n$ goes to infinity, but the harmonic series is known to diverge, so the sequence does not converge.

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  • $\begingroup$ shouldn't |$x_{n+1} - x_n $| be 1/(n+1) - 1? $\endgroup$ Dec 16, 2013 at 1:51

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