大学数学 级数

证明收敛且求其和
<1>n/(2n-1)^2*(2n+1)^2
<2>1/n(n+1)(n+2)
已邀请:

donkeycn

赞同来自: 张新月 Math001 数分学弟

(1)$\sum_{1}^{+∞}$$\cfrac{n}{({2n-1})^{2}(2n+1)^2}$
=$\cfrac{1}{8}$$\sum_{1}^{+∞}$$(\cfrac{1}{({2n-1})^2}-\cfrac{1}{({2n+1})^2})$
=$\cfrac{1}{8}$$((1-\cfrac{1}{9})+(\cfrac{1}{9}-\cfrac{1}{25})+(\cfrac{1}{25}-\cfrac{1}{49})+......)$
=$\cfrac{1}{8}$

(2)$\sum_{1}^{+∞}$$\cfrac{1}{n(n+1)(n+2)}$
=$\cfrac{1}{2}$$\sum_{1}^{+∞}$$\cfrac{(n+2)-n}{n(n+1)(n+2)}$
=$\cfrac{1}{2}$$\sum_{1}^{+∞}$$(\cfrac{1}{n(n+1)}-\cfrac{1}{(n+1)(n+2)})$
=$\cfrac{1}{2}$$\cdot$$\cfrac{1}{2}$
=$\cfrac{1}{4}$

要回复问题请先登录注册