$\sum\sin(\frac{k\pi}{4})$ absolute convergent, conditional convergent,divergent? Would the following be abs convergent, conditional convergent or divergent. 
$\sum\sin(\frac{k\pi}{4})$
I know sin(x) is between $-1<x<1$
y=sin(x) is oscillating would it be $(-1)^n$

Sorry about the old answer, that was just garbage.
The real answer is that it's not convergent. You can try grouping them in terms of $8$ like I had done, but the problem is that, because it ends at $\infty$, you don't have a point to end at. That means the answer could be the same as any of the following eight:
$\displaystyle \sum_{k=1}^1 \sin \left(\frac{k \pi}{4}\right) = \frac{\sqrt{2}}{2}$.
$\displaystyle \sum_{k=1}^2 \sin \left(\frac{k \pi}{4}\right) = \frac{\sqrt{2}}{2} + 1$.
$\displaystyle \sum_{k=1}^3 \sin \left(\frac{k \pi}{4}\right) = \frac{\sqrt{2}}{2}$.
$\displaystyle \sum_{k=1}^4 \sin \left(\frac{k \pi}{4}\right) = 0$.
$\displaystyle \sum_{k=1}^5 \sin \left(\frac{k \pi}{4}\right) = -\frac{\sqrt{2}}{2}$.
$\displaystyle \sum_{k=1}^6 \sin \left(\frac{k \pi}{4}\right) = -\frac{\sqrt{2}}{2} - 1$.
$\displaystyle \sum_{k=1}^7 \sin \left(\frac{k \pi}{4}\right) = -\frac{\sqrt{2}}{2}$.
$\displaystyle \sum_{k=1}^8 \sin \left(\frac{k \pi}{4}\right) = 0$.
Because they're different, and because you have no sure way of determining which is the right answer, the answer is that the series does not converge.

This is actually a classic problem of trying to find the sum of an alternating series. Let's take a similar problem, where you're trying to find the answer to $$1 - 1 + 1 - 1 + \cdots$$ (This is called Grandi's Series, btw.)
There are multiple answers, depending on how you group them. You can group them as $(1-1) + (1-1) + \cdots = 0 + 0 + \cdots = 0$, as $1 + (-1 + 1) + (-1 + 1) + \cdots = 1 + 0 + 0 + \cdots = 1$, or even as $\displaystyle \frac{1}{2} + \left(\frac{1}{2} - \frac{1}{2}\right) + \left(-\frac{1}{2} + \frac{1}{2}\right) + \cdots = \frac{1}{2} + 0 + 0 + \cdots = \frac{1}{2}$. Since there's no "right" way of doing this, the answer is DNE. Hence, it does not converge.
