Does this sequence have an $n$th term? $-1,0,1,0,-1,0,1...$ So my sequence is actually
$$-\frac16, 0, \frac{1}{120}, 0, -\frac{1}{5040}, 0, \frac{1}{362880}$$
which can be simplified to
$$-1/3! \ , \ \ 0\ , \ \ 1/5!\ , \ \ 0, \ \ -1/7!\ , \ \ 0\ , \ \ 1/9!$$
I've figured out the denominator would be $(n+3)!$ (My sequence starts with $n=0$). 
I've tried a lot of ways with $(-1)^n$ etc. to get the negatives and positives right but can't figure that out, nor how to get the zeroes in the nth term.
Any help would be appreciated, thanks.
 A: Hint: use trig functions $\sin(x)$ and/or $\cos(x)$. Adjust their period using $ k\pi x$ where $k$ is some real number.
(if you don't want to mess with complex numbers) 
A: If you want to avoid trig functions and imaginary numbers, try
$$(-1)^{\lfloor n/2\rfloor+1}\left(1+(-1)^n\over2\right)$$
where $\lfloor\cdot\rfloor$ is the greatest integer (aka "floor") function. This gives the sequence $-1,0,1,0,-1,0,1,0,\ldots$ starting with $n=0$.
Alternatively, if you want to avoid the greatest integer function as well, this'll do:
$$-(-1)^{n(n+1)/2}\left(1+(-1)^n\over2\right)$$
The expression $n(n+1)/2$ gives $0,1,3,6,10,15,21,\ldots$, which, for even $n$ alternates between even and odd.
A: Consider
$$\frac{(-1)^{\frac{n(n+1)}2} + (-1)^{\frac{(n+1)(n+2)}2}}2$$
The idea is that $\frac{n(n+1)}2$ has the pattern $$\text{odd}, \text{odd}, \text{even}, \text{even}$$
and $\frac{(n+1)(n+2)}2$ has the pattern$$\text{odd}, \text{even}, \text{even}, \text{odd}.$$
A: This sequence may be given by $$i^{n-1}\frac{(-1)^n-1}{2}.$$ Hope this helps!
