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

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.

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