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