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I'm looking for a sequence $(a_n)_{n∈\mathbb{N}+}$ which does not converge but $a_n^2-4a_n$ converges to 0. I tried looking at trigonometric functions and functions of the form $(-1)^n$, but none of them seem to fit the criteria. Any help/hints would be appreciated.

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  • $\begingroup$ For what values of $x$ is $x^2 - 4x = 0$? $\endgroup$ Oct 30, 2016 at 17:27

2 Answers 2

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Let $a_{2n}=0$ and $a_{2n+1}=4$.

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HINT: Notice that $a_n^2-4a_n=a_n(a_n-4)$; can you find a sequence that takes exactly two values in such a way that $a_n^2-4a_n$ converges?

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  • $\begingroup$ The limit of $a_n=0$ and the limit of $a_n-4=0$? But in that case $a_n$ converges to $0$. Am I missing something? $\endgroup$
    – aL_eX
    Oct 30, 2016 at 17:32
  • $\begingroup$ @aL_eX: No, make the sequence alternate between two different values. $\endgroup$ Oct 30, 2016 at 17:33
  • $\begingroup$ The only way I can think of doing that is by using $a_n=(-1)^n$ $\endgroup$
    – aL_eX
    Oct 30, 2016 at 17:35
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    $\begingroup$ @aL_eX: You don’t have to have a formula! If $c$ and $d$ are the two values in question, you can, for instance, let $a_n=c$ if $n$ is even and $a_n=d$ if $n$ is odd. $\endgroup$ Oct 30, 2016 at 17:36
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    $\begingroup$ @aL_eX: Yep, you’ve got it. You’re welcome! $\endgroup$ Oct 30, 2016 at 17:44

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