# Lim sup, lim inf definition/convention

1. May 11, 2007

### 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.

2. May 11, 2007

### NateTG

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.

3. May 11, 2007

### quasar987

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

4. May 11, 2007

### HallsofIvy

Staff Emeritus
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= $\infty$.

5. May 11, 2007

### 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.

6. May 11, 2007

### AKG

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

7. May 11, 2007

No, no.

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