Calc 2: convergent of divergent sequences I would like to know if this sequence, $\sin\left(\frac{n \pi}{2}\right)$ ,is convergent or divergent? 
I have done this problem and I know that it is divergent through oscillation (I'm pretty sure). Is there a mathematical way to prove this? If so can you please show me. Thank you  
 A: If a sequence is convergent, then its limit is unique, and every subsequence converges to that limit.
Can you find two subsequences of your sequence that have a different limit?
A: Let $k$ be an integer, $|\sin((2k)\pi/2) - \sin((2k + 1)\pi/2)| = 1$. Thus when $\epsilon = 1$, for all $N$ we can find elements in the sequence which differ by $\epsilon$. Let $k = N$. This contradicts the definition of convergence. 
A: In the field of real analysis, you can show that every convergent sequence $\left\{a_n\right\}_n$ is what we call a Cauchy sequence: that is, for every $\epsilon > 0$, we can find an integer $N$ so that for all pairs of integers $m,n > N$, 
$$|a_m - a_n| < \epsilon.$$
(In fact, every Cauchy sequence also converges, but we won't need this information to show that your sequence diverges.)
To see this from a more intuitive angle, imagine your favorite convergent sequence.  Let's call it $\left\{a_n\right\}_n$, and let's say it converges to some number we'll call $A$.  Then, when $n$ is really big, $a_n \approx A$.  So, if both $n$ and $m$ are both really big, $a_n - a_m \approx A - A = 0$.
Now, let's return to your sequence $\left\{a_n\right\}_n = \left\{\sin\left(\frac{n\pi}{2}\right)\right\}_n = \left\{1, 0, -1, 0, 1, 0, -1, \cdots\right\}_n$.  Then, for any $n$
$$\left|a_n - a_{n+1}\right| = 1.$$
So the terms of your sequence are not getting closer together (it is not a Cauchy sequence), and thus it must diverge.
