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