1. The series converges absolutely.

    1. Take f(x)= x/ex, so an= (-1)ⁿf(n). Note that f'(x)= (1 – x)/e^x, so f(x) is decreasing for x ≥ 1, and           by l’Hopital’s rule lim_{x → ∞}x/e^x= 0. Then by the Alternating Series Test, the series converges.           To test if the convergence is conditional or absolute consider the series b_n= |a_n|. Apply the         Limit Comparison Test to b_n and 1/n₂: (n/eⁿ)/(1/n²)= n³/eⁿ → 0 as n → ∞ To see the last,                 replace n with x and apply l’Hopital’s rule three times. Then ∑b_n converges by the Limit                 Comparison Test and so ∑an converges absolutely.

    1. The series converges conditionally.

    1. Let f(x)= cos(1/x)/x, so a_n= (-1)_nf(n). Note f'(x)= (-xcos(1/x) + sin(1/x))/x³, negative for x >                   1.163, so f(n) is decreasing for n > 1. Now lim_{x → ∞}cos(1/x)/x = 0, so by the Alternating Series         Test the series ∑a_n converges. To test if the convergence is conditional or absolute consider         the series b_n= |a_n|. The series ∑b_n diverges by the applying the Limit Comparison Test to             the harmonic series. Then the series ∑a_n converges conditionally.

    1. The series does not converge absolutely.

    1. If the series converged absolutely, then by the rearrangement theorem, every                                 rearrangement would converge to the same limit. But an obvious rearrangement of this               series is the alternating harmonic series, which does not converge absolutely.                                 Consequently, the original series does not converge absolutely.