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.

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Answer 2

The series Σ((-1)^n)/((2n+1)!) from n=1 to infinity converges absolutely because the absolute value of each term is bounded by a convergent series, namely Σ(1/((2n+1)!)).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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