Determine whether the series $\sum _{n=1}^{\infty \:\:}\frac{\left(-1\right)^n}{(3n)!}$ is convergent or divergent.

Question

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?

Answer 1

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.