Question

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

Answer 1

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.