For what values of $\epsilon > 0$ is it nonetheless true that... The sequence $s_n = (-1)^n$ does not converge. For what values of $\epsilon > 0$ is it nonetheless true that there is an integer N so that $|s_n-1|< \epsilon $ whenever $ n \geq N$? For what values of $\epsilon > 0$ is it nonetheless true that there is an integer N so that $|s_n-0|< \epsilon $ whenever $ n \geq N$? (Pictures can help.) I have so far that $\epsilon = 2$ for part one (if that is even right) but am having trouble understanding the intuition and carry over to the second part.

Note that if $n$ is even then $s_n = 1$. If so then $|s_n - 1| = |1 -1| = 0$. And if $n$ is odd then $s_n = -1$. If so then $|s_n - 1| = |-1-1|= 2$. So $|s_n - 1| = 0$ or $2$ and $|s_n - 1| \le 2$. So as long as $\epsilon > 2$ it will be true that $|s_n - 1| < \epsilon$. This is true for _all_ $n$ and not just "significantly large" $n$, so we don't have to worry about there being an $N$ that $n$ has to be larger than.

Now by the same reasoning, only easier, $|s_n - 0| = |s_n|$ and if $n$ is even $s_n = 1$ and if $n$ is odd $s_n = -1$. So $|s_n -0| = |s_n| = |\pm 1| = 1\le 1$. So for any $\epsilon > 1$ we will always have $|s_n -0| < \epsilon$. Again this is true for _all_ $n$; not just $n > N$ for some $N$.