幂级数(-1)^n∠1⼀n+1是绝对收敛还是条件收敛

幂级数(-1)^n∠1/n+1是绝对收敛还是条件收敛
2025-03-14 21:34:14
推荐回答(3个)
回答1:

条件收敛.

(1)因为|(-1)^n/(n+1)|=1/(n+1),而∑1/(n+1)发散,所以∑|(-1)^n/(n+1)|发散;

(2)因为1/(n+1)单调递减且lim(n—>无穷)1/(n+1)=0,所以由Leibniz交错级数判别法知∑(-1)^n/(n+1)收敛.

综上,级数条件收敛.

扩展资料

条件收敛

一般的级数u1+u2+...+un+...

它的各项为任意级数。

如果级数Σu各项的绝对值所构成的正项级数Σ∣un∣收敛,

则称级数Σun绝对收敛。

如果级数Σun收敛,

而Σ∣un∣发散,

则称级数Σun条件收敛。

回答2:

条件收敛。

分析过程如下:

(1)因为|(-1)^n/(n+1)|=1/(n+1),而∑1/(n+1)发散,所以∑|(-1)^n/(n+1)|发散;

(2)因为1/(n+1)单调递减且lim(n—>无穷)1/(n+1)=0,所以由Leibniz交错级数判别法知∑(-1)^n/(n+1)收敛。

综上,幂级数(-1)^n•1/n+1条件收敛。

扩展资料:

绝对收敛一般用来描述无穷级数或无穷积分的收敛情况。如果级数ΣUn各项的绝对值所构成的级数Σ|Un|收敛,则称级数ΣUn绝对收敛,级数ΣUn称为绝对收敛级数。绝对收敛级数一定收敛。

若函数f(x)在[a,b]上可积,且|f(x)|的无穷积分(从a到+∞)上收敛,则称 f(x) 的无穷积分(从a到+∞)绝对收敛。绝对收敛一定收敛。

由条件收敛级数重排后所得的新级数,即使收敛,也不一定收敛于原来的和数。而且,条件收敛级数适当排列后,可得到发散级数,或收敛于事先任意指定的数。

回答3:

条件收敛。
(1)因为|(-1)^n/(n+1)|=1/(n+1),而∑1/(n+1)发散,所以∑|(-1)^n/(n+1)|发散;
(2)因为1/(n+1)单调递减且lim(n—>无穷)1/(n+1)=0,所以由Leibniz交错级数判别法知∑(-1)^n/(n+1)收敛。
综上,级数条件收敛。

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