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.

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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.

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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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