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Given is the following sequence $a_n=(-n)^{(-n)^n}$ $(n \in \mathbb{N})$. Find all limit points.

Here's what I have so far, I divided it in three cases.

Case 1: $-n > 0 \rightarrow n < 0$

I did this "trick" $m=-n, a_n=m^{m^{-m}}$

$m^{-m}$ converges to $0$ and therefore, $a = m^0 = 1$, when $n$ tends to infinity. This is our only limit point.

Case 2: $-n < 0 \rightarrow n > 0$

I divided into two more cases here:

n is even: $-n^n$ diverges to negative infinity, therefore $a_n=(-n)^{(-n)^n}$ also. n is odd: $-n^n$ diverges to positive infinity, therefore $a_n=(-n)^{(-n)^n}$ also.

Is this correct? Thanks

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Just because $g(x)$ converges to $0$ doesn't mean that $f(x)^{g(x)}$ converges to 1. So that fact that $m^{-m}\to 0$ doesn't mean that $m^{m^{-m}}\to 1$ – Thomas Andrews Nov 15 '11 at 19:02
Case 2: $n$ even: $-n^n\rightarrow-\infty$; but $(-n)^n$ does not... – David Mitra Nov 15 '11 at 19:03
1. what is the domain of $n$, doesn't it take only positive values? 2. which exponentiation goes first? 3. as @Thomas has pointed out, your argumentation is incorrect, as a hint you may think about using logarithms – Ilya Nov 15 '11 at 19:03
In case1: $n\to\infty\Rightarrow m\to-infty$. – Tapu Nov 15 '11 at 19:04
0 should be one (look at the odd positive integers) – David Mitra Nov 15 '11 at 19:13
up vote 2 down vote accepted

From the wording I’d have expected the $a_n$ to be defined only for $n>0$ (or possibly for $n\ge 0$). Assume that $n>0$ is even, say $n=2m$. Then $(-n)^n=(-2m)^{2m}=(4m^2)^m$, which clearly is even and increases without bound as $m$ does, and $a_n=a_{2m}$ does the same.

Now assume that $n$ is odd, say $n=2m+1$. Then $(-n)^n=(-2m-1)^{2m+1}=-|n|^{|n|}$, so $a_n=\frac1{|n|^{|n|^{|n|}}}$, which clearly approaches $0$ as $n$ increases without bound.

If $n<0$, let $m=-n$, so that $a_n=m^{m^{-m}}$, and take logs: $\ln a_{-m}=m^{-m}\ln m=\dfrac{\ln m}{m^m}$. Since $m^m=e^{m\ln m}$, L’Hospital’s rule yields $$\lim_{n\to\infty}\ln a_{-m} = \lim_{n\to\infty}\frac{1/m}{1+\ln m}=0\;,$$ whence $\lim\limits_{n\to -\infty}a_n = 1$.

Thus, the sequence $\langle a_n:n\in\mathbb{Z}^+\rangle$ has no cluster points in $\mathbb{R}$; in the extended reals both $\infty$ and $-\infty$ are cluster points. The bisequence $\langle a_n:n\in\mathbb{Z}\rangle$ has one real cluster point, $1$, and the cluster points $\infty$ and $-\infty$ in the extended reals.

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sorry Brian, I guess you saw my question before I edited it. The domain is $\mathbb N$, as you had predicted – Clash Nov 15 '11 at 19:39
@Clash: I did indeed. Oh well; a little extra information can’t hurt. – Brian M. Scott Nov 15 '11 at 19:41
@BrianM.Scott +1 for "extra information" :) – Tapu Nov 15 '11 at 19:44

Here's a correct argument.

If $n$ is even, then $a_n$ has the form $(-n)^{k}$, where $k=n^n$ is even; whence $\lim\limits_{n\rm\ even} a_n=\infty$.

If $n$ is odd, then $a_n$ has the form ${1\over (-n)^{k}}$, where $k=n^n$ ; whence $\lim\limits_{n\rm\ odd} a_n=0$.

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Thanks for your reply! Why is $k=n^n$ and not $k=-n^n$ for $n$ is even? I do understand for $n$ is odd, as you now have a division. But for even I didn't get it... thanks in advance again! – Clash Nov 15 '11 at 19:30
because you are taking $(-n)^n$ (note the parentheses). If $n$ is even, the negative sign is "killed". For $n$ odd, $(-n)^n=-n^n$, and you can bring $(-n)^{-n^n}$ "downstairs" if you drop the negative sign. – David Mitra Nov 15 '11 at 19:34

Well, your question was first unclear to me because of "all limit points" in the title (though I have edited your question, I did not change this part). Please note the following:

  • Any sequence can converge to at most one limit.
  • A sequence is convergent if and only if all of its sub-sequence converges to the same limit (of the original sequence).

So, in your case the sub-sequence $\{a_{2n}\}$ diverges (as you have noted). So, the sequence $\{a_{n}\}$ also diverges.

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a "limit point" is just some terminology, see – Tyler Nov 15 '11 at 19:29
Clearly limit point is being used here in the sense cluster points or accumulation point; a sequence may have many of these. E.g., a sequence that enumerates the rationals has every real as a cluster point. – Brian M. Scott Nov 15 '11 at 19:37
@TylerBailey It looks I have messed up with "limit" and "limit-points". – Tapu Nov 15 '11 at 19:38

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