Finding set of all accumilation values -- how can I be sure that I have found all? I am having some trouble in my analysis class with Subsequences and accumulation values.
Consider the series:
$\{x_n\}_n=(-1)^n+\frac{1}{n}$.
I need to find the set off all accumilation values.
To do this I am trying to find all convergent subsequences and then I calulate their limit.
Now I know that $x_{2n}=1+\frac{1}{2k}$ and $x_{2n-1}=-1+\frac{1}{2k-1}$ converge to 1 and -1 respectivly so I would argue that the set of all accumilation values is $\{-1,1\}$.
But how can I be sure that those are the only convergent subseqences, and to that extent, the only accumilation values?
I have just been trying to find more accumilation values with brute force and cannot, so I am assuming that there are none left for me to find.
Is there a rigourous mathematical arguement I can make to be sure that there are no more extra accumilation values that I am missing???
If so, how to I apply this arguement to future problems with different series so I do not get stuck on the same snag again.
Many thanks in advance.
 A: It looks like you've got all the pieces, and just need to put them together!
You've already noticed (and should be able to prove) the following:

*

*Given any real $\epsilon_0>0,$ there is some natural $N_0$ such that for all $n\ge N_0,$ we have that $n$ is odd or $|x_n-1|<\epsilon_0.$

*Given any real $\epsilon_1>0,$ there is some natural $N_1$ such that for all $n\ge N_1,$ we have that $n$ is even or $|x_n+1|<\epsilon_1.$
From the above, it immediately follows that for any real $\epsilon>0,$ there are infinitely-many $n$ such that $|x_n-1|<\epsilon$ and $|x_n+1|<\epsilon,$ meaning that $1$ and $-1$ are accumulation points of the given sequence, but we can do even better, and prove the following:

*

*Given any real $\epsilon>0,$ there is some natural $N$ such that for all $n\ge N,$ we have that $|x_n-1|<\epsilon$ or $|x_n+1|<\epsilon.$
From that result, we can directly prove the following:

*

*Given any real $x$ such that $x\neq\pm 1,$ there is some natural $N$ such that for all $n\ge N,$ we have $|x_n-x|\ge|x-1|$ and $|x_n-x|\ge|x+1|.$
Finally, by choosing an appropriate $\epsilon>0$ to correspond with our arbitrary $x\neq\pm 1,$ we conclude from the previous result that there are at most finitely-many $n$ for which $|x_n-x|<\epsilon,$ and so $x$ cannot be a point of accumulation of the given sequence.
Can you prove each of the first four results, and see what choice of $\epsilon$ will work to prove the desired conclusion?
A: If $x$ is a limit (ie accumulation) point of $\{x_n\}_n$ then, $\ \forall\ \varepsilon:\quad \forall\ M,\ \exists\ N>M$ such that $\ \vert x_N - x\vert < \varepsilon.$
The contrapositive to the above statement is:
If $\ \forall\ \varepsilon:\ \exists\ M$ such that $\ \not\exists\ N>M\ $ with $\ \vert x_N-x \vert < \varepsilon,\ $ then $x$ is not a limit point of $\{x_n\}_n.$
Or, to put it another way:
If $\ \forall\ \varepsilon:\ \exists\ M$ such that $\ N>M\implies \vert x_N-x \vert \geq \varepsilon,\ $ then $x$ is not a limit point of $\{x_n\}_n.\quad (1)$
We suspect that $-1$ and $1$ are the only limit points of $\{x_n\}_n.$ But to prove this, suppose $x\neq \pm1,$ and let $\varepsilon>0.$ Show that an $M$ as in statement $(1)$ exists.
A: This is my 'device' to tackle this problem very easily:
Theorem

*

*Let $(i_n),(j_n),(k_n)$ are increasing sequences of natural numbers

*Let  $$I:=\{i_n\,|\,n\in\Bbb N\},\ J=\{j_n\,|\,n\in\Bbb N\},\ K:=\{k_n\,|\,n\in\Bbb N\}.$$
Assume $I\cup J \cup K \supset \Bbb N_\alpha=\{n\in\Bbb N\,|\,n\geq \alpha\}$ for some $\alpha$.

*Assume $x_{i_n}\to a$, $x_{j_n}\to b$, $x_{k_n}\to c$.

Then the the set of accumulation points is $\{a,b,c\}$.
Remark
This also works for two or any finite number of increasing sequences of natural numbers.
For example, consider $\{x_n\}_n=(-1)^n+\frac{1}{n}$.
Then:

*

*Define $i_n = 2n$, $j_n=2n+1$.

*Observe that $\{i_n\,|\,n\in\Bbb N\}\cup \{j_n\,|\,n\in\Bbb N\} = \Bbb N_2$ (each number is even even or odd).

*$x_{i_n}=1+1/n\to 1$, $x_{j_n}=-1+1/n\to -1$.

Then $\pm 1$ are the only two accumulation points.
Edit (after being downvoted because of no proof):
Sketch of the proof (of the theorem).
Of course $a,b,c$ are accumulation points. Now consider any acc point $d=\lim_{n\to\infty}x_{m_n}$. Let $M:=\{m_n:n\in\Bbb N\}$. Then $M\cap \Bbb N_\alpha$ is infinite and
$$M\cap \Bbb N_\alpha \subset (M\cap I)\cup(M\cap J)\cup(M\cap K).$$ Therefore  (from the Pigeonhole principle for infinite sets) at least one of these sets if infinite, say $\#M\cap I=\infty$, so
$$M\cap I = \{m_{s_1},m_{s_2},\ldots\}.$$ On one hand,
$x_{m_{s_n}}$ is a subsequence of $x_{m_n}$, on the other, it's a subsequence of $x_{i_n}$. Therefore
$$ d\leftarrow x_{m_{s_n}}\to a.$$ We get $d=a$, so we got no new acc points.
