Calculus of a Single Variable, Chapter 9, 9.5, Section 9.5, Problem 19

We may apply the Ratio Test to determine the convergence or divergence of the series sum_(n=0)^oo (-1)^n/(n!) .
In Ratio test, we determine the limit as:
lim_(n-gtoo)|a_(n+1)/a_n| = L
Then, we follow the conditions:
a) L lt1 then the series is absolutely convergent
b) Lgt1 then the series is divergent.
c) L=1 or does not exist then the test is inconclusive. The series may be divergent, conditionally convergent, or absolutely convergent.

For the series sum_(n=0)^oo (-1)^n/(n!) , we have a_n=(-1)^n/(n!) .
Then, we may let a_(n+1) =(-1)^(n+1)/((n+1)!)
We set up the limit as:
lim_(n-gtoo) |((-1)^(n+1)/((n+1)!)) /((-1)^n/(n!))|
To simplify the function, we flip the bottom and proceed to multiplication:
|((-1)^(n+1)/((n+1)!)) /((-1)^n/(n!))|=|(-1)^(n+1)/((n+1)!) * (n!)/(-1)^n|
Apply Law of Exponent: x^(n+m) = x^n*x^m and (n+1)! = n!(n+1)
|((-1)^n(-1)^1)/(n!(n+1)) * (n!)/(-1)^n|
Cancel out the common factors (-1)^n and n! .
|(-1)^1/(n+1)|
=|-1/(n+1)|
=1/(n+1)
Applying |((-1)^(n+1)/((n+1)!)) /((-1)^n/(n!))|=1/(n+1) , we get:
lim_(n-gtoo) |((-1)^(n+1)/((n+1)!)) /((-1)^n/(n!))|
=lim_(n-gtoo)1/(n+1)
=(lim_(n-gtoo)1)/(lim_(n-gtoo)(n+1))
= 1 /oo
= 0
The limit value L=0 satisfies the condition: L lt1 .
Therefore, the series sum_(n=0)^oo (-1)^n/(n!) is absolutely convergent.