How Can the Alternating Series Test Assume N=1?

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SUMMARY

The discussion centers on the Alternating Series Test, which requires that all terms \( u_n \) are positive, that \( u_n \geq u_{n+1} \) for all \( n \geq N \), and that \( u_n \) approaches 0. A participant questions the assumption of \( N=1 \) in the theorem, arguing that the sum of a finite number of terms cannot equal infinity. The conversation emphasizes the validity of the test for arbitrary \( N \) and encourages readers to explore proofs by reducing to the case of \( N=1.

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georg gill
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Alternating series test:

1. All the u_n are all positive

2. u_n\geq u_{n+1} for all n \geq N. For some integer N

3. u_n \rightarrow 0


I thought it would hold with 2. and that the su m of the N first terms were not \infty

Here is the theroem just in case:

http://bildr.no/view/1047382

Here they assume N=1 how can they do that?
 
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georg gill said:
Alternating series test:

1. All the u_n are all positive

2. u_n\geq u_{n+1} for all n \geq N. For some integer N

3. u_n \rightarrow 0


I thought it would hold with 2. and that the su m of the N first terms were not \infty

How can the sum of the first N terms ever equal infinity?? If you add up finitely many real number, then you never get infinity.

Here is the theroem just in case:

http://bildr.no/view/1047382

Here they assume N=1 how can they do that?

The proof for arbitrary N is very similar. Try to prove it yourself!
(or even better: reduce to the case N=1)
 

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