`sum_(n=0)^oo (-1)^n/(n!)` Determine the convergence or divergence of the series.

Expert Answers
marizi eNotes educator| Certified Educator

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.