Accumulation Points for $S = \{(-1)^n + \frac1n \mid n \in \mathbb{N}\}$ I was recently asked to find the accumulation points of the set $$S = \{(-1)^n + \frac{1}{n} \mid n \in \mathbb{N}\}$$
I answered that the accumulation points are $\{-1,1\}$, because despite the fact that $\frac{1}{n}$ diverges, we can still use $\frac{1}{n}$ to find an element arbitrarily close to either $1$ or $-1$, i.e. to show that there is an element in an arbitrarily small neighborhood of $1$ or $-1$. (No proof was required for this question --- this is my intuition.)
Am I right at all?

Let's denote
$$a_n=(-1)^n+\frac{1}{n}$$
then the subsequence $(a_{2n})$ and $(a_{2n+1})$ are convergent to $1$ and $-1$ respectively then $1$ and $-1$ are two accumulation points. There's not other accumulation point since any convergent subsequence of $((-1)^n)$ has either $1$ or $-1$ as limit.
