利用逐项积分或逐项求导，求下列级数的和函数：(1)∑n=1∞nxn-1(2)∑n=1∞xn/n(3)∑n=0∞x2n+1/(2n
2024-08-03高等数学（工本）(00023)
利用逐项积分或逐项求导，求下列级数的和函数：
(1)∑n=1∞nxn-1
(2)∑n=1∞xn/n
(3)∑n=0∞x2n+1/(2n+1).
【正确答案】：(1)∑n=1∞nxn-1=(∑n=1∞xn)′ =[(x/(1-x)]′=1/(1-x)2 (-1﹤x﹤1) (2)∵(∑n=1∞xn/n)′=∑n=1∞xn=1= (1/x)∑n=1∞xn=1/x•[x/(1-x)]=1/(1-x) ∴∑n=1∞xn/n=∫[1/(1-x)]dx=-ln(1-x) (-1≤x≤1) (3)∵[∑n=0∞x2n+1/(2n+1)]′=∑n=0∞x2n= x2/(1-x2)2 ∴∑n=0∞x2n+1/(2n+1)=∫[x2/(1-x2)2]dx= (1/2)ln[(1+x)/(1-x)] (-1﹤x﹤1)
(1)∑n=1∞nxn-1
(2)∑n=1∞xn/n
(3)∑n=0∞x2n+1/(2n+1).
【正确答案】：(1)∑n=1∞nxn-1=(∑n=1∞xn)′ =[(x/(1-x)]′=1/(1-x)2 (-1﹤x﹤1) (2)∵(∑n=1∞xn/n)′=∑n=1∞xn=1= (1/x)∑n=1∞xn=1/x•[x/(1-x)]=1/(1-x) ∴∑n=1∞xn/n=∫[1/(1-x)]dx=-ln(1-x) (-1≤x≤1) (3)∵[∑n=0∞x2n+1/(2n+1)]′=∑n=0∞x2n= x2/(1-x2)2 ∴∑n=0∞x2n+1/(2n+1)=∫[x2/(1-x2)2]dx= (1/2)ln[(1+x)/(1-x)] (-1﹤x﹤1)
