$$\sum _{ n=1 }^{ \infty }{ \frac { (-4)^ n }{ 3^{2n+2} } } $$
I am confused whether to use alternating series test or the ratio test.
When I used ratio test then it converges. My doubt is that since the series alternates, we should use alternating series test .
On
Use the root or ratio test. Note that the series $\sum ar^n$ converges iff $|r|<1$ so even if $r$ is negative and it alternates the geometric property is so strong that it overrides all other concerns.
Just to be clear this particular series IS geometric so you use the criterion for convergence of a geometric series.
On
The ratio test is fine, assuming you applied absolute values when using the ratio test, since the ratio test requires the terms to be strictly positive.
This works because absolute convergence implies convergence.
However, as a point of technicality, we have an alternating geometric series here, so if you apply the ratio test with absolute values, you get a geometric series.
You cannot use the ratio test to prove that a geometric series converges.
This is because the proof of the ratio test requires the fact that a geometric series converges, so we would have a circular argument.
You can use a bunch of different tests, to get a variety of answers - each of them correct, but not necessarily equally informative.
In order from worst to best:
The alternating series test. This says that a series converges if (not only if!) the terms go to $0$, alternate sign, and have decreasing absolute value. The $n$th term in the series is $$a_n=(-1)^n\cdot ({4\over 9})^n\cdot{1\over 9},$$ so the alternating series test tells us that this converges.
The root test. While the alternating series test told us that the series converges, it didn't tell us whether it converged absolutely or merely conditionally. If we apply the root test, we will figure this out: we have $$\lim_{n\rightarrow\infty}\sqrt[n]{\vert a_n\vert}=\lim_{n\rightarrow\infty}{4\over 9}\cdot {1\over\sqrt[n]{9}}=0,$$
so the series converges absolutely.
The ratio test. We have $\lim_{n\rightarrow\infty}{a_n\over a_{n+1}}=-{9\over 4}$, so - just like the root test - the ratio test says that the series converges absolutely. Like the root test, this provides more information than the alternating series test (which just said that the series converged).
But think about what we just did: the ratio of successive terms doesn't just approach $-{9\over 4}$, it always is ${9\over 4}$. That is, we're dealing with a geometric series whose ratio is $-{4\over 9}$ (which is strictly between $-1$ and $1$). This tells us not only that the series converges absolutely, but $\color{red}{\mbox{the exact value it converges to}}$ - namely, $${a_1\over 1-r}={-{4\over 81}\over {13\over 9}}=-{4\over 117}.$$ This is the right thing to do in this case. It gives the most information, and is as easy if not easier than anything else you can do here.
(Incidentally, the only reason I ranked the ratio test better than the root test is that it points out that the series is geomteric, if one has not already noticed that.)