document.write( "Question 1176853: Does the series |-1 + 1/e^n| converges or conditionally converges where n is from 0 to infinity? \n" ); document.write( "
Algebra.Com's Answer #850560 by CPhill(1959)\"\" \"About 
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Let's analyze the convergence of the series Σ | -1 + 1/e^n | where n ranges from 0 to infinity.\r
\n" ); document.write( "\n" ); document.write( "**1. Analyze the Terms**\r
\n" ); document.write( "\n" ); document.write( "* The terms of the series are given by | -1 + 1/e^n |.
\n" ); document.write( "* As n approaches infinity, 1/e^n approaches 0.
\n" ); document.write( "* Therefore, as n becomes large, the terms | -1 + 1/e^n | approach | -1 + 0 | = 1.\r
\n" ); document.write( "\n" ); document.write( "**2. Apply the Divergence Test**\r
\n" ); document.write( "\n" ); document.write( "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.\r
\n" ); document.write( "\n" ); document.write( "* lim (n→∞) | -1 + 1/e^n | = 1\r
\n" ); document.write( "\n" ); document.write( "Since the limit of the terms is 1 (not 0), the series diverges by the Divergence Test.\r
\n" ); document.write( "\n" ); document.write( "**3. Conclusion**\r
\n" ); document.write( "\n" ); document.write( "The series Σ | -1 + 1/e^n | diverges. It does not converge, and therefore it cannot conditionally converge.\r
\n" ); document.write( "\n" ); document.write( "**Note on the Code Output**\r
\n" ); document.write( "\n" ); document.write( "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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