How do you determine if the series the converges conditionally, absolutely or diverges given #Sigma ((-1)^(n))/((2n+1)!)# from #[1,oo)#?

1 Answer
MoominDave
Jun 5, 2018

Compare with known convergent series #Sigma_(n=1)^oo1/(n!)#. The Maclaurin series for #e^x# is #Sigma_(n=0)^oox^n/(n!)#, convergent #AA x in RR#, so #e=Sigma_(n=0)^oo1/(n!)#. Chop one term from the front: #Sigma_(n=1)^oo1/(n!)# converges, to the value #e-1#.

To test the series in the question for absolute convergence, sum the absolute value of each term;
#Sigma_(n=1)^oo1/((2n+1)!)#

We see that this series of uniformly positive terms has a set of terms that is a subset of the set for the series of uniformly positive terms compared to, which converges to a known value.

Hence the series is absolutely convergent.

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