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.
Well the limit is zero, so you need to prove that is a strictly decreasing sequence.
Put $\displaystyle f(x)=\frac x{e^x},$ so $\displaystyle f'(x)=\frac{{{e}^{x}}-x{{e}^{x}}}{{{e}^{2x}}}=\frac{{{e}^{x}}(1-x)}{{{e}^{2x}}}<0,$ for $\displaystyle x\ge2.$
By the alternating series test, your series converges.