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If I calculate a limit and get the value $\infty$, what is the proper way to communicate this? Can I say that the $\lim_{n\to\infty}a_n=\infty$ and therefore the sequence $\{a_n\}$ diverges, or do I say that the limit does not exist?

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Both are common, so the answer most likely depends on how much of a pedant your instructor is. –  MJD May 13 '13 at 15:19
    
If using the extended real number system, the answer is completely unambiguous, namely it 'converges to $\infty$'. –  DepeHb May 13 '13 at 15:22

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I would say the limit "tends to" $\infty$ and therefore your sequence diverges.

When working out limits, what you're doing is looking at what happens as you get very close to some point. So when working out the limit as $n \to \infty$, you're looking to see what happens as $n$ gets very big and as you're not working (whatever) out at a specific point, I wouldn't say its correct to say it "equals". "Tends to" means that it gets close to that point.

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Saying that the limit is infinity has the same precise meaning as writing $\lim_{n\to\infty}a_n=\infty$. Saying that the sequence diverges has not the same precise meaning, because the limit could also be $-\infty$.

I'd avoid phrases such as "the limit tends to"; the limit either doesn't exist or, if it exists, it is a real number, $\infty$ or $-\infty$.

In any case, it's a "local" question: terminology is not fixed; somebody uses "divergent" as synonymous to "non convergent", for instance. But what you're allowed to write you should be also allowed to say.

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Saying it diverges could also mean that there simply isn't a limit, finite or not. For instance $(-1)^n$. –  DepeHb May 13 '13 at 15:23
    
@DepeHb It depends on conventions. In the courses I took, "divergent" had always the meaning I said. For $(-1)^n$ we used "oscillating". –  egreg May 13 '13 at 15:25

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