Finding all limit points of the sequence $a_n=(-n)^{(-n)^n}$ 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
 A: 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 $\lim\limits_{n\rm\  odd} a_n=0$.
A: 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.
A: 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.
