sum_(n=1)^oo 1/(n!) Use the Root Test to determine the convergence or divergence of the series.

It is usually easier to use ratio test on these types of series that contain factorials. However, we can also use root test if we rewrite factorial using exponentials. This can be accomplished using Stirling's approximation
n! approx sqrt(2pi n)(n/e)^n
The reason why we can use this approximation is because it becomes more precise for greater values of n, in fact the ratio of the left and right hand side of the approximation converges to 1 as n tends to infinity.
lim_(n to infty)root(n)(1/(n!))=lim_(n to infty)1/root(n)(n!)=
Now we use Stirling's approximation.
lim_(n to infty)1/root(n)(sqrt(2pi n)(n/e)^n)=
In order to calculate lim_(n to infty)root(n)(sqrt(2pi n)) we need to use the following two facts:
lim_(n to infty) root(n)(c)=1,  c in RR and lim_(n to infty)root(n)(n^p)=1,  p in RR.
                                                                                                                    Applying this to our limit yields
lim_(n to infty)1/root(n)(sqrt(2pi n)(n/e)^n)=lim_(n to infty)1/(n/e)=lim_(n to infty)e/n=e/infty=0
Since the value of the limit is less than 1, according to the root test, the series is convergent. 
https://en.wikipedia.org/wiki/Root_test

https://en.wikipedia.org/wiki/Stirling%27s_approximation