1. ## Leibniz Test

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.

2. Well the limit is zero, so you need to prove that is a strictly decreasing sequence.

Put $f(x)=\frac x{e^x},$ so $f'(x)=\frac{{{e}^{x}}-x{{e}^{x}}}{{{e}^{2x}}}=\frac{{{e}^{x}}(1-x)}{{{e}^{2x}}}<0,$ for $x\ge2.$

By the alternating series test, your series converges.

3. Could I plug n+1 for all n as well to prove that its decreasing?

n+1/e^(n+1) < n/e^(n)

4. 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.