SOLUTION: Does the series |-1 + 1/e^n| converges or conditionally converges where n is from 0 to infinity?

Question 1176853: Does the series |-1 + 1/e^n| converges or conditionally converges where n is from 0 to infinity?
Answer by CPhill(1959) (Show Source): You can put this solution on YOUR website!
Let's analyze the convergence of the series Σ | -1 + 1/e^n | where n ranges from 0 to infinity.
**1. Analyze the Terms**
* The terms of the series are given by | -1 + 1/e^n |.
* As n approaches infinity, 1/e^n approaches 0.
* Therefore, as n becomes large, the terms | -1 + 1/e^n | approach | -1 + 0 | = 1.
**2. Apply the Divergence Test**
The Divergence Test states that if the limit of the terms of a series does not approach zero as n approaches infinity, then the series diverges.
* lim (n→∞) | -1 + 1/e^n | = 1
Since the limit of the terms is 1 (not 0), the series diverges by the Divergence Test.
**3. Conclusion**
The series Σ | -1 + 1/e^n | diverges. It does not converge, and therefore it cannot conditionally converge.
**Note on the Code Output**
The provided code calculates the partial sum of the series up to a certain number of terms. However, it does not determine the true convergence of the infinite series. The code's output "The series converges" is incorrect because the series actually diverges.

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