What are good techniques for identifying formulas in sequences? I've been having some trouble identifying the formulas for some of the more difficult sequences in my textbook. 
Most sequences I've seen come in fractions, which is nice to do. I always try and identify some kind of pattern between each numerator and each denominator, and find a relation between them and $n$. It's worked well, but some of these more difficult ones are getting the better of me. For instance:
$1: \{-3, 2, -\frac43, \frac89, -\frac{16}{27},...\}$
Normally, when I see $-,+,-,+,..$ I assume a negative base, with $n$ somewhere in the exponent, generally because that allows it to switch between positive and negative. Here, you see that happening, which is a good  start. I also notice how the numerator from $n=2$ continues to double, and how the denominator continues to triple from $n=3$, but the $-3$ and the $-2$ completely throw that off. 
Another thing to note, is that there is a possiblility of this formula being in the format $\{a_{n+1}\}_{n=1}^{infinity}$. I usually see this when the first few terms are whole numbers that are previously defined, and then suddenly throws itself into fractions, so that's also a possibility. 
Other than those few clues, I'm completely thrown off, especially by the jump between $n=2$ and $n=3$.
Help with this question is appreciated, but I would also appreciate any tips on ways to go about identifying patterns in sequences. I know there isn't a formula way, but if there any tactics that are considered useful and worth noting, I would definitely love to hear them.
Much thanks
 A: Three general techniques to help one find simple patterns in sequences:


*

*Compute the forward differences $\Delta a_n:= a_{n+1}-a_n$. Might need to do twice.

*Compute the successive ratios $r_n:=a_{n+1}/a_n$ in case these are simpler.

*Find the prime factorizations of the terms. See if the exponents of primes exhibit patterns.


You do not have to do all of the above, nor do you have to do any one of these fully; a quick scan-over should suffice to tell us that powers of two and three are appearing increasingly in the numerator and denominator respectively, and the signs are alternating consistently. This is enough to suggest a common ratio; compute $r_1=-2/3$ and check that indeed each successive term is the prior one multiplied by $-\frac{2}{3}$. Our sequence will thus be of the form $a_n=a_1 r^{n-1}=-3(-2/3)^{n-1}$.
The reason the first two terms were integers while the others fractions is because the first term was contained in the common ratio's denominator. Don't let something like a couple numbers being whole-valued throw you off the trail; check the possibilities that jump out at you sufficiently.
