What are the limit points of the sequence ${(-1)^n+\frac{2}{n}}$? I know that 1 and -1 are the limit points. But how do i prove these are the only ones (If they are, but i am pretty sure they are) Do i have to show that every sequence made from the set only converge to these numbers? How would i do that?
 A: Look at the quantity $x_n - x_m$, where $x_j = (-1)^j + \frac{2}{j}$.
It is equal to $(-1)^n- (-1)^m + \frac{2}{n} - \frac{2}{m}$.
Any convergent sequence is Cauchy, so we want to see what kind of subsequences of $x_n$ can possibly be Cauchy.
Note that $\frac{2}{n} - \frac{2}{m}$ would go to zero anyway, so it doesn't affect the Cauchy property of any sequence. On the other hand, $$
|(-1)^n - (-1)^m| = \begin{cases}
0 & n - m \text{ even}\\
1 & n - m \text{ odd}
\end{cases}
$$ 
Hence, we want $n$ and $m$ to eventually have the same parity, in order to give convergence,  otherwise it will not happen.

Hence a subsequence $\{ x_{n_k}\}$ of $x_n$ is convergent if and only if $n_k$ eventually has the same parity (even/odd).

Now, it is easy to see  that if $n_k$ has odd parity, then $(-1)^n = -1$, so the sequence $x_{n_k}$ converges to $-1$.
Similarly, it is easy to see  that if $n_k$ has even parity, then $(-1)^n = 1$, so the sequence $x_{n_k}$ converges to $1$.
Hence, the only other limit points are $1$ and $-1$.
