Determine whether the series $\sum _{n=1}^{\infty \:\:}\frac{\left(-1\right)^n}{(3n)!}$ is convergent or divergent. Determine whether the series
$\sum _{n=1}^{\infty \:\:}\frac{\left(-1\right)^n}{(3n)!}$ 
is convergent or divergent. If it is convergent, then how many terms of the series do we need in order to find the sum to within $10^{-5}$? 
I'm trying to use the Alternating Series Test
$a_n=\left(-1\right)^n\:$
$b_n=\frac{1}{\left(3n\right)!}$
need to check if 
$\lim _{n\to \infty }\left(\frac{1}{\left(3n\right)!}\right)$ $=\:0$
which is does because 
$\frac{1->\:1}{\left(3n\right)!\:->\infty \:}\:=0$
then, need to check if $\left\{\frac{1}{\left(3n\right)!}\right\}$ is decreasing
$b_{n+1}\frac{1}{\left(3\left(n+1\right)\right)!}$ $=\:\frac{1}{(3n+3)!}$
$b_{n+1}<\:b_n$ and therefore decreasing
so the series is convergent by Alternating Series Test
Now for the other part of the question start writing out the series
$\sum _{n=1}^{\infty }\:\frac{\left(-1\right)^n}{\left(3n\right)!}\:=\:\frac{-1}{3!}+\frac{1}{6!}-\frac{1}{9!}\:$
$\frac{1}{6!}=1.388\cdot 10^{-3}$
$-\frac{1}{9!}\:=-2.755\:\cdot 10^{-6}$
since the question is within $10^{-5}$ i would think not to include $-\frac{1}{9!}$
and say i only include 2 terms. Does my work and answers look correct?

Edit: This is an answer for the previous version, where an error was $10^{-15}$. According to a suggestion I am leaving it, as a hint, what one should do if the error is smaller than $10^{-5}$.
Near good, but in your question is $10^{-15}$. It is known, that the error is less than the firts of remaining terms, hence $(3n)!>10^{15}$, which is true from $n=6$, hence you need two additional terms.
