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Can I say that if I know that a sequence is of the following form:
$$\lim_{n \to \infty} a_n \ne L$$
Then one can know for sure that $a_n$ is divergent?

So proving that $\lim_{n \to \infty} a_n \ne L$ will actually prove that $a_n$ is divergent

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  • $\begingroup$ The expression $\lim_{n \to \infty} a_n$ is only defined if the sequence $(a_n)$ is convergent. $\lim_{n \to \infty} a_n \ne L$ means that the sequence is convergent and the limit is not $L$. Your question is unclear to me. $\endgroup$ – Martin R Nov 15 '14 at 17:06
  • $\begingroup$ @MartinR I think Im looking for a way to prove that a sequence is divergent or more precisely finding the mathematical definition $\endgroup$ – The One Nov 15 '14 at 17:21
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What about $a_n = (-1)^n$?

The point is that the limit may not exists

So, if you prove that the limit exists, and it cannot be equal to any real number, than it must be $\pm \infty$ (Of course I'm assuming your sequence to be real-valued)

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  • $\begingroup$ So how could I prove that a sequence is divergent? $\endgroup$ – The One Nov 15 '14 at 17:20
  • $\begingroup$ @kfir124 proving that a sequence is divergent is proving that $\lim _{n \to \infty} a_n = \infty$. Usually one can use the usual rules to do that easily (it is not difficult to see that $a_n = n$ is divergent).. Or if you want to be formal, you can make a $\epsilon -\delta $ proof following the definition of limit. $\endgroup$ – Ant Nov 15 '14 at 17:24

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