Determine whether the sequence $\sin\frac{n\pi}{2}$ converges and prove your answer using the epsilon method.. I am having the most trouble with the scratch work ahead of the actual proof. I can't figure out proof of $x_n=\sin\frac{n\pi}{2}$ beyond $\epsilon \in \mathbb{R}$ and $\epsilon > 0$.
 A: HINT:
Note that the subsequence $x_{2n}\equiv 0$, the subsequence $x_{4n+1}\equiv1$, and the subsequence $x_{4n+3}\equiv-1$.  
What can you conclude now?
A: It might be helpful to consider the actual values of $\sin(n\pi / 2)$ where $n \in \mathbb{N}$ to obtain an alternate definition of the sequence.
For example, try $n = 1, 2, 3, 4,...$ and so on, and see if you can find a pattern.  This seems more like a trick question because of that: once you get the other, equivalent sequence, things should be much easier. 
(You might have to properly justify that the two sequences are equivalent though but that should be easy enough.)
A: If the series converges to some real number $l$, then $\forall\varepsilon>0, \exists n_0\in\mathbb N$ such that $|\sin \frac{n\pi}{2}-l|<\varepsilon, \forall n\geq n_0$. It suffices to show that $\exists\varepsilon>0$ for which no such $n_0$ exists.
Say $\varepsilon=1/10$. The interval $(l-\varepsilon, l+\varepsilon)\equiv(l-1/10, l+1/10)$ has length 1/5. Say $\sin \frac{n\pi}{2}\in(l-1/10, l+1/10)$
$\sin \frac{(n+2)\pi}{2}=-\sin\frac{n\pi}{2}\in(l-1/10, l+1/10)$
or, by the definition of an interval, $(-\sin\frac{n\pi}{2}, \sin \frac{n\pi}{2})\subset(l-1/10, l+1/10)$. Whatever be $n_0$ in this case, you can always find a number of the form $n=4m+3\geq n_0, m\in\mathbb N$. 
$(-\sin\frac{n\pi}{2}, \sin \frac{n\pi}{2})\equiv(-1,1)$
What is the length of this interval?
A: Let $$a_n=x_{2n}=\sin n\pi =0\\b_n=x_{{4n+1}}=\sin {4n+1\over 2}\pi=1$$since $a_n$ and $b_n$ are convergent to different limits, therefore $x_n$ is not convergent.
