Limit points of the set $A=\lbrace(-1)^{n}(1+\frac{1}{n})^{n-1}\mid n\in\mathbb{N}\rbrace$ I have searched for a question similar to this and unfortunately I have not found one.Let $A=\lbrace(-1)^{n}(1+\frac{1}{n})^{n-1}\mid n\in\mathbb{N}\rbrace$ be a subset of Euclidean space $(\mathbb{R},|.|)$. If we denote the set of limit points of $A$ with $A'$, then is it true that $A'=\emptyset$?I must insist that the proof should only contain the concepts regarding limit points and neighborhoods and I am not allowed to use any sequence related concepts. I tried to introduce $0<r_{0}\in\mathbb{R}$ where $N_{r_{0}}(x)\cap A=\emptyset$ for any $x\in\mathbb{R}$ where $r_{0}=\frac{1}{2}(|(-1)^{n}(1+\frac{1}{n})^{n-1}-(-1)^{n-1}(1+\frac{1}{n})^{n-2}|)=\frac{1}{2}(2+\frac{1}{n})(1+\frac{1}{n})^{n-2}\\$ but I cannot proceed any further. Any helps are most appreciated.

Maybe it would be useful to make a partition of $A$ according to the parity of $n$ and then consider the supremum/infimum of the resulting sets.
$A=A^+ \cup A^-$, where $A^+=\{e^{(2n-1)\log (1+1/2n)}, n\in \mathbb{N} \} $ and $A^-=\{-e^{2n\log (1+1/(2n+1))}, n\in \mathbb{N} \} $.
It remains to study the limit points of $\{n \log(1+1/n)\}$,which is unique: 1
