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Suppose that $(a_n)$ is a sequence of integers. Prove that $(a_n)$ converges if and only if there is a natural number $N$ big enough and an integer $m$ such that for any $n \geq N$, ${a_n} = m$. I thought this is the definition of convergence of a sequence. Anyone can guide me on how to prove this ?

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    $\begingroup$ The phrase "for any $n\geq N$, $\ \lim_n a_n=m\ $" makes no sense. $\endgroup$ – Christian Blatter Feb 3 '13 at 15:39
  • $\begingroup$ @ChristianBlatter: mind to elaborate ? $\endgroup$ – Idonknow Feb 3 '13 at 15:43
  • $\begingroup$ The letter $n$ in $\lim_{n\to\infty} a_n$ is burnt up in the process; therefore it has no meaning outside of the $\lim$. Note that $\lim_{n\to\infty} a_n$ is the same as $\lim_{k\to\infty} a_k$. $\endgroup$ – Christian Blatter Feb 3 '13 at 15:49
  • $\begingroup$ The $\lim a_n = m$ should just be $a_n = m$, right? $\endgroup$ – John Martin Feb 3 '13 at 15:52
  • $\begingroup$ Reread your definition of cenvergence. If you still come up with this, discard your sources. $\endgroup$ – vonbrand Feb 3 '13 at 19:37
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Take any $0<\epsilon<\frac12$, and use the fact that convergent sequences are Cauchy to conclude that a convergent sequence of integers must be (eventually) constant. The other direction should be trivial.

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  • $\begingroup$ for the other direction, I just have to quote the definition of limit and say the sequence converges ? $\endgroup$ – Idonknow Feb 3 '13 at 15:48
  • $\begingroup$ Once you remove that erroneous "$\lim$" from the given statement, that's basically all you have to do. $\endgroup$ – Cameron Buie Feb 3 '13 at 15:52
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A sequence $(a_n)$ of integers converges if and only of there is a natural number $N$ and an integer $m$ such that $a_n=m$ for all $n\ge N$. This follows from the fact that the (usual) topology on the set of integers is discrete.

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  • $\begingroup$ But the question now asks us to prove the statement without the knowledge of topology. This is a real analysis question $\endgroup$ – Idonknow Feb 3 '13 at 15:33
  • $\begingroup$ Is it typo ? $a_n=m \,\,\,\forall n \geq N$? $\endgroup$ – Idonknow Feb 3 '13 at 15:39
  • $\begingroup$ @Idonknow: No, that is correct. There shouldn't be a "$\lim$" there. $\endgroup$ – Cameron Buie Feb 3 '13 at 15:44
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Assume that $\lim_{n\to\infty} a_n=\xi\in\Bbb R$. There is an $N\in\Bbb N$ such that $a_n\in I:=\ \bigl]\xi-{1\over3},\xi+{1\over3}\bigr[\ $ for all $n>N$. As $I$ contains at most one point $m\in\Bbb Z$ it follows that $a_n=m$ for all $n>N$. $-$ The converse is obvious.

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That is wrong. Consider the sequence $a_n = 1 - 2^{n + 1}$, which converges to $1$. No integer $a_n$ in sight, no constant value for $a_n$ starting at some $N$ whatsoever.

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    $\begingroup$ I think you meant something else, but note that $(a_n)$ is stated in the question to be a sequence of integers. Your sequence (which is likely a typo) does not converge. If you meant $1-2^{-n}$, then it is not a sequence of integers. $\endgroup$ – Jonas Meyer Feb 4 '13 at 2:23

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