Question 1176853
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.