Calculus of a Single Variable, Chapter 9, 9.5, Section 9.5, Problem 39

To determine the convergence or divergence of the series sum_(n=1)^oo (-1)^n/(n!) , we may apply the Ratio Test.
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 converges absolutely
b) Lgt1 then the series diverges
c) L=1 or does not exist then the test is inconclusive.The series may be divergent, conditionally convergent, or absolutely convergent.
For the given series sum_(n=1)^oo (-1)^n/(n!) , we have a_n =(-1)^n/(n!) .
Then, 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 . It becomes:
| ((-1)^n (-1)^1)/((n+1)!)*(n!)/(-1)^n|
Cancel out common factors (-1)^n and apply (-1)^1 = -1
| -(n!)/((n+1)!) |
Simplify:
| -(n!)/((n+1)!) |=(n!)/((n+1)!)
=(n!)/(n!(n+1))
=1/(n+1)
The limit becomes:
lim_(n-gtoo)1/(n+1) =(lim_(n-gtoo) (1))/(lim_(n-gtoo) (n+1) )
= 1/(oo+1)
=1/oo
=0
The limit value L=0 satisfies the condition: L lt1 .
Therefore, the series sum_(n=1)^oo (-1)^n/(n!) is absolutely convergent.