• anonymous
Does the series converge absolutely, converge conditionally, or diverge? Sum ((-1)^n)*(n!/(e^(n^2)), n-> 1 to infinity. I would like to confirm my answer. Since -1^n is a alternating series, let |asubn| = n!/e^(n^2). Then using the ratio test, let the limit as n-> infinity equal |asubn+1/asubn|. Which in this case will come out to be the absolute value |(n+1)/(e^(2n+1))| letting n->infinity become x->infinity, we find: Limit as x->infinity (x+1)/(e^(2x+1)). this is Infinity/infinity which can be converted to (1)/(2e^(2x+1)) which equals 0. Therefore L = 0 and thus the L < 1.
Mathematics
• Stacey Warren - Expert brainly.com
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SOLVED
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