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

Question

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.

anonymous · Answer

Great work. What is your conclusion?