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

1 Answer
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.