I am trying determine the set of values of $x \in \mathbb{R}$ for which the following series converges.
$$ \sum_{n=1}^\infty{\frac{(-1)^nx^{3n}}{n!}} $$
But I am yet to understand how to work with these types of series. For what I may understand (if I interpreter correctly) is to determine whether the modulus values of the sum increases or decreases, find a upper or lower limit and the character of that limit would determine that of the sum. Is that correct?
How do I proceed?
On
$${\displaystyle e^{x}=\sum _{n=0}^{\infty }{\frac {x^{n}}{n!}}=1+x+{\frac {x^{2}}{2!}}+{\frac {x^{3}}{3!}}+\cdots }$$ $${\displaystyle e^{x}-1=\sum _{n=1}^{\infty }{\frac {x^{n}}{n!}}}$$ now let $x=-x^3$ to get $${\displaystyle e^{-x^3}-1=\sum _{n=1}^{\infty }{\frac {-x^{3n}}{n!}}}=\sum _{n=1}^{\infty }{\frac {(-1)^nx^{3n}}{n!}}$$ that means it converges for all values of $x$
$$ \left|\frac{\left(-1\right)^nx^{3n}}{n!}\right|=\frac{\left|x\right|^{3n}}{n!} $$
Then you can use the d'Alembert critera to show it converges for all $x \in \mathbb{R}$.