Infinity over Sigma n=1 ((-1)^n)*(n)*(e^-n) Now I know Im suppose to use the leibniz test and I got as far as lanl=n/e^n but Im not sure where to go from there.
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Well the limit is zero, so you need to prove that is a strictly decreasing sequence.
Put so for
By the alternating series test, your series converges.
Could I plug n+1 for all n as well to prove that its decreasing?
n+1/e^(n+1) < n/e^(n)
Yes but you need to prove that such inequality holds.
The best and fast way to prove that general term is a strictly decreasing sequence is to contemplate its derivative as I did above.
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