首先有幂级数展开:1/(1-x) = ∑{0 ≤ n} x^n.
求导得1/(1-x)² = ∑{0 ≤ n} nx^(n-1) = ∑{1 ≤ n} nx^(n-1).
于是x/(1-x)² = ∑{1 ≤ n} nx^n.
代入x = 1/3得3/4 = ∑{1 ≤ n} n/3^n.
也有别的办法.
例如设a = ∑{1 ≤ n} n/3^n,则a/3 = ∑{1 ≤ n} n/3^(n+1)
= ∑{1 ≤ n} (n+1)/3^(n+1)-∑{1 ≤ n} 1/3^(n+1)
= ∑{2 ≤ n} n/3^n-1/9·1/(1-1/3)
= a-1/3-1/6
= a-1/2.
解得a = 3/4.