分类：教育/科学-理工学科-数学
问题：求级数∞∑n=1n/[（n+1）（n+2）（n+3）]的和答案是 
答案是2/9呀但是我算的不是呀
回答：n=1→∞，求：∑{n/[(n+1)(n+2)(n+3)]}

an = (n+2)/[(n+1)(n+2)(n+3)]
　 = 1/[(n+1)(n+3)]
　 = (1/2)[(n+3)-(n+1)]/[(n+1)(n+3)]
　 = (1/2)[1/(n+1)-1/(n+3)]
--->a1+a2+a3+a4+... = (1/2)[1/2+1/3] = 5/12

bn = 2/[(n+1)(n+2)(n+3)]
　 = (1/2)[(n+3)-(n+1)]/[(n+1)(n+2)(n+3)]
　 = (1/2){1/[(n+1)(n+2)]-1/[(n+2)(n+3)]}
--->b1+b2+b3+b4+... = (1/2)[1/(2•3)] = 1/12

∑{n/[(n+1)(n+2)(n+3)]} = ∑an - ∑bn = 5/12-1/12 = 1/4