Need help finding the limit of geometric serires I'm learning some series tests in calculus and I can't completely figure this out. I know it's easier than i'm making it. Here's the question:

Determine whether the geometric series is divergent or convergent. If
it is convergent, evaluate its limit. If it diverges state your answer
as DIV.
$2/3-(2/3)^3+(2/3)^5-(2/3)^7$

I know that it's convergent. But i'm unsure of how to find the limit. Do I have to find the common ratio between them?
My teacher is saying I have to find the "$n$-th term" first. Which is just $(2/3)^n$
 A: You have a difference between the sum to infinity of two geometric series, both with ratio $\left(\frac{2}{3}\right)^4$:-
$$\sum_{k=0}^\infty\left(\frac{2}{3}\right)^{4k+1}-\sum_{k=0}^\infty\left(\frac{2}{3}\right)^{4k+3}$$ 
For a geometric series with first term $a$ and ratio $r$, the sum to infinity is given by formula $$S_\infty=\frac{a}{1-r}$$
Thus we have
$$\sum_{k=0}^\infty\left(\frac{2}{3}\right)^{4k+1}-\sum_{k=0}^\infty\left(\frac{2}{3}\right)^{4k+3}=\frac{\frac{2}{3}}{1-\left(\frac{2}{3}\right)^4}-\frac{\left(\frac{2}{3}\right)^3}{1-\left(\frac{2}{3}\right)^4}=\frac{6}{13}$$ 
A: Your $n$-th term is
$$\left(\frac{2}{3}\right)^{2n+1}(-1)^n=\frac{2}{3}\left(-\frac{4}{9}\right)^{n}$$
So your series is
$$\frac{2}{3}\sum_{n=0}^{\infty}\left(-\frac{4}{9}\right)^{n}$$
If you know the formula
$$\sum_{n=0}^{\infty}q^n=\frac{1}{1-q}$$
then you're done.
A: The two numbers you need for an geometric series are the first term, $a$, and the common ratio, $r$.
If $|r|<1$ then the sum to infinity is given by the formula
$$S_{\infty} = \frac{a}{1-r}$$
Clearly $a=\frac{2}{3}$ in your example, but what is the common ratio $r$? What number do you need to multiply each term to get the next term?
Once you have $r$, use the formula for $S_{\infty}$.
I suspect you've been asked to find the $n^{\text{th}}$-term to use the ratio test to prove convergence. A geometric series looks like $a+ar+ar^2+ar^3+\cdots$ and so the $n^{\text{th}}$-term is $U_n = ar^{n-1}$.
A: In your case, applying the formula for the summation of geometric series, we see that the sum of the first $n$ terms is $q \frac{-q^{2n}-1}{-q^2-1}$ (where $q= \frac{2}{3}$) That's because the first term is $q$ and the denumenator is $-q^2$. Now observe that $q^2n$ approaches zero, therefore the limit is $\frac{q}{q^2+1}$.
