{n/a^n}的求和怎么做

问题描述:

{n/a^n}的求和怎么做

设起和为S
S=∑n/a^n=1/a + 2/a^2 + 3/a^3 + ...
aS=1 + 2/a + 3/a^2 + 4/a^3 + ...
-:
(a-1)S=1 + 1/a + 1/a^2 + 1/a^3 + ...
=1/[1-(1/a)]
=a/(a-1)
所以S=a/[(a-1)^2]