Lim sup, lim inf definition/convention

[quasar987]

My book says that if the set of all cluster points is empty, then we write lim sup = $-\infty$ and if the sequence is not bounded above, we write limsup = $+\infty$.


But what if both happen at the same time? for instance consider x_n=1/n. There are no accumulation points and it is unbounded above.

 

[NateTG]

But what if both happen at the same time? for instance consider x_n=1/n. There are no accumulation points and it is unbounded above.

Maybe I'm getting too tired, but
$$1>\frac{1}{n}$$
is an upper bound, and
$$0$$
is an accumulation point.

You might consider $x_n=n(-1)^n$ which is unbounded in the reals, and doesn't have any real accumulation points. However, in the extedended reals, $\pm \infty$ are cluster points of that sequence, and $+ \infty$ is an upper bound for all sequences.

 

[quasar987]

It is I who is too tired. I meant to say x_n = n.

 

[HallsofIvy]

My book says that if the set of all cluster points is empty, then we write lim sup = $-\infty$ and if the sequence is not bounded above, we write limsup = $+\infty$.


But what if both happen at the same time? for instance consider x_n= n. There are no accumulation points and it is unbounded above.

I don't see a problem. Since the set of all cluster points is empty, lim sup= $-\infty$ and since the sequence is not bounded above, limsup= $\infty$.

 

[quasar987]

Well, actually, the author give the same x_n=n as an example and he write lim sup=+$\infty$.

As if the fact that it is not bounded above takes priority over the fact that the set of all cluster points is empty.

 

[AKG]

Are "lim sup" and "limsup" supposed to be two different things?

 

[quasar987]

No, no.

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