Question

求幂级数∞∑n=1 [x^( 2n-1) ]/ ( 2n-1) 的收敛区间与函数，并求数项级数∞∑n=1 [n(n+1) ]/ ( 2^n)的和

需要详细的解题步骤

Answer 1

lim(n→∞) |[x^(2n+1)/(2n+1)]/[(x^(2n-1)/(2n-1)]
=x²<1
则收敛半径为1
当x=1时，该级数为[∞∑n=1]1/(2n-1) 发散
当x=-1时，该级数为[∞∑n=1]-1/(2n-1)发散
∴收敛区间为(-1,1)
[∞∑n=1] [x^( 2n-1) ]/ ( 2n-1)
=[∞∑n=1 ]∫x^( 2n-2) dx (积分区间为0到x)
=∫[∞∑n=1 ]x^( 2n-2) dx (积分区间为0到x)
=∫[1/(1-x²)]dx (积分区间为0到x)
=1/2ln|(1+x)/(1-x)|

构造一个幂级数[∞∑n=1 ]n(n+1) x^n
=x[∞∑n=1 ]n(n+1) x^(n-1)
=x[∞∑n=1 ][(n+1) x^n]′
=x[[∞∑n=1 ](n+1) x^n]′
[∞∑n=1 ](n+1) x^n
=[∞∑n=1 ][x^(n+1)]′
=[[∞∑n=1 ]x^(n+1)]′
=[x²/(1-x)]′
=x(2-x)/(1-x)²
该幂级数=x[x(2-x)/(1-x)²]′
=2x/(1-x)³
将x=1/2代入该级数得
∞∑n=1 [n(n+1) ]/ ( 2^n)=8