求下列级数的和函数:∑(n=0,∞)x^n⼀2^n

2024-11-23 10:28:20
推荐回答(2个)
回答1:

解:
∑(n=0,∞)x^n/2^n
=1+x/2+(x/2)^2+(x/2)^3+……+(x/2)^n
收敛域为(-1,1)
∑(n=0,∞)x^n/2^n
=1+x/2+(x/2)^2+(x/2)^3+……+(x/2)^n即等比数列
=1+[(x/2)(1-(x/2)^n)]/(1-(x/2))
=1+x/2(-1

回答2:

因为:要∑(n=0,∞)x^n/2^n收敛,x<2,∑(n=0,n)x^k/2^k=[1-(x/2)^(n+1)]/(1-x/2)
:∑(n=0,∞)x^n/2^n=1/(1-x/2)