u_n = Σ_(k=n+1) ^(2n) (1/k) and v_n = Σ_(k=n) ^(2n−1) (1/k) • show that u_n and v_n are adjacent use ln(x+1) ≤ x and x≤−ln(1−x) and • show that u_n ≤ Σ_(k=n+1) ^(2n) (ln(k)−ln(k−1)) hence deduce that u_n ≤ ln2 • show that v_n ≥ Σ_(k=n) ^(2n−1) (ln(k+1)−ln(k)) hence deduce that v_n ≥ln2