Student Question

`sum_(n=1)^oo (-1)^n/(n!)` Determine whether the series converges absolutely or conditionally, or diverges.

Expert Answers

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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)!) |`


`| -(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.

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