求下列函数展开成x-1的幂级数,并求其收敛域 ln(x+2)

网友回答

令t=x-1
则x=t+1
ln(x+2)
=ln(t+3)
=ln3+ln(1+t/3)
由ln(1+x)=x-x²/2+x^3/3-,收敛域-1
======以下答案可供参考======
供参考答案1:
1/(x+2)=1/[3+(x-1)]=(1/3)/[1+(x-1)/3] = (1/3)∑(-1)^n[(x-1)/3]^n
= ∑(-1)^n(x-1)^n/3^(n+1).
则 ln(x+2)-ln3 =∫dt/(t+2)
=∫ ∑(-1)^n(t-1)^n/3^(n+1)dt
= ∑[(-1)^n/3^(n+1)]∫(t-1)^nd(t-1)
= ∑[(-1)^n/3^(n+1)][(x-1)^(n+1)/(n+1)],
得 ln(x+2) = ln3 +∑[(-1)^n/3^(n+1)][(x-1)^(n+1)/(n+1)]。
收敛域 -1