求和：n/2^n （n=1,2,3,…,∞）

不十分严谨但是结果正确的简单求法如下:
1/2+2/4+3/8+4/16+5/32+...
= 1/2+1/4+1/8+1/16+1/32+...
+1/4+1/8+1/16+1/32+...
+1/8+1/16+1/32+...
+1/16+1/32+...
+1/32+...
...
= 1+1/2+1/4+1/8+1/16+...
= 2.
一种严谨的做法如下:
首先由D'Alembert比值判别法易得∑{1 ≤ n} n/2^n收敛, 设S = ∑{1 ≤ n} n/2^n.
则S/2 = ∑{1 ≤ n} n/2^(n+1)
= ∑{1 ≤ n} (n+1)/2^(n+1) - ∑{1 ≤ n} 1/2^(n+1)
= ∑{2 ≤ n} n/2^n -1/2
= ∑{1 ≤ n} n/2^n -1/2-1/2
= S-1.
解得S = 2.