【课后习题】高等数学第七版下第十二章无穷级数第二节常数项级数的审敛法

习题12-2

1. 用比较审敛法或极限形式的比较审敛法判定下列级数的收敛性:

(1) 1+13+15+⋯+1(2n−1)+⋯1+\frac{1}{3}+\frac{1}{5}+\cdots+\frac{1}{(2 n-1)}+\cdots1+31+51+⋯+(2n−1)1+⋯;

(2) 1+1+21+22+1+31+32+⋯+1+n1+n2+⋯1+\frac{1+2}{1+2^2}+\frac{1+3}{1+3^2}+\cdots+\frac{1+n}{1+n^2}+\cdots1+1+221+2+1+321+3+⋯+1+n21+n+⋯;

(3) 12⋅5+13⋅6+⋯+1(n+1)(n+4)+⋯\frac{1}{2 \cdot 5}+\frac{1}{3 \cdot 6}+\cdots+\frac{1}{(n+1)(n+4)}+\cdots2⋅51+3⋅61+⋯+(n+1)(n+4)1+⋯;

(4) sin⁡π2+sin⁡π22+sin⁡π23+⋯+sin⁡π2n+⋯\sin \frac{\pi}{2}+\sin \frac{\pi}{2^2}+\sin \frac{\pi}{2^3}+\cdots+\sin \frac{\pi}{2^n}+\cdotssin2π+sin22π+sin23π+⋯+sin2nπ+⋯;

(5) ∑n=1∞11+an(a>0)\sum_{n=1}^{\infty} \frac{1}{1+a^n} \quad(a>0)∑n=1∞1+an1(a>0).

2. 用比值审敛法判定下列级数的收敛性:

(1) 31⋅2+322⋅22+333⋅23+⋯+3nn⋅2n+⋯\frac{3}{1 \cdot 2}+\frac{3^2}{2 \cdot 2^2}+\frac{3^3}{3 \cdot 2^3}+\cdots+\frac{3^n}{n \cdot 2^n}+\cdots1⋅23+2⋅2232+3⋅2333+⋯+n⋅2n3n+⋯;

(2) ∑n=1∞n23n\sum_{n=1}^{\infty} \frac{n^2}{3^n}∑n=1∞3nn2;

(3) ∑n=1∞2n⋅n!nn\sum_{n=1}^{\infty} \frac{2^n \cdot n !}{n^n}∑n=1∞nn2n⋅n!;

(4) ∑n=1∞ntan⁡π2n+1\sum_{n=1}^{\infty} n \tan \frac{\pi}{2^{n+1}}∑n=1∞ntan2n+1π.

3. 用根值审敛法判定下列级数的收敛性:

(1) ∑n=1∞(n2n+1)n\sum_{n=1}^{\infty}\left(\frac{n}{2 n+1}\right)^n∑n=1∞(2n+1n)n;

(2) ∑n=1∞1[ln⁡(n+1)]n\sum_{n=1}^{\infty} \frac{1}{[\ln (n+1)]^n}∑n=1∞[ln(n+1)]n1;

(3) ∑n=1∞(n3n−1)2n−1\sum_{n=1}^{\infty}\left(\frac{n}{3 n-1}\right)^{2 n-1}∑n=1∞(3n−1n)2n−1;

(4) ∑n=1∞(ban)n\sum_{n=1}^{\infty}\left(\frac{b}{a_n}\right)^n∑n=1∞(anb)n, 其中 an→a(n→∞),an,b,aa_n \rightarrow a(n \rightarrow \infty), a_n, b, aan→a(n→∞),an,b,a 均为正数.

4. 判定下列级数的收敛性 :

(1) 34+2(34)2+3(34)3+⋯+n(34)n+⋯\frac{3}{4}+2\left(\frac{3}{4}\right)^2+3\left(\frac{3}{4}\right)^3+\cdots+n\left(\frac{3}{4}\right)^n+\cdots43+2(43)2+3(43)3+⋯+n(43)n+⋯;

(2) 141!+242!+343!+⋯+n4n!+⋯\frac{1^4}{1 !}+\frac{2^4}{2 !}+\frac{3^4}{3 !}+\cdots+\frac{n^4}{n !}+\cdots1!14+2!24+3!34+⋯+n!n4+⋯

(3) ∑n=1∞n+1n(n+2)\sum_{n=1}^{\infty} \frac{n+1}{n(n+2)}∑n=1∞n(n+2)n+1;

(4) ∑n=1∞2nsin⁡π3n\sum_{n=1}^{\infty} 2^n \sin \frac{\pi}{3^n}∑n=1∞2nsin3nπ;

(5) 2+32+⋯+n+1n+⋯\sqrt{2}+\sqrt{\frac{3}{2}}+\cdots+\sqrt{\frac{n+1}{n}}+\cdots2+23+⋯+nn+1+⋯;

(6) 1a+b+12a+b+⋯+1na+b+⋯(a>0,b>0)\frac{1}{a+b}+\frac{1}{2 a+b}+\cdots+\frac{1}{n a+b}+\cdots \quad(a>0, b>0)a+b1+2a+b1+⋯+na+b1+⋯(a>0,b>0).

5. 判定下列级数是否收敛? 如果是收敛的, 是绝对收敛还是条件收敛?

(1) 1−12+13−14+⋯+(−1)n−1n+⋯1-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{4}}+\cdots+\frac{(-1)^{n-1}}{\sqrt{n}}+\cdots1−21+31−41+⋯+n(−1)n−1+⋯;

(2) ∑n=1∞(−1)n−1n3n−1\sum_{n=1}^{\infty}(-1)^{n-1} \frac{n}{3^{n-1}}∑n=1∞(−1)n−13n−1n;

(3) 13⋅12−13⋅122+13⋅123−13⋅124+⋯+(−1)n−113⋅12n+⋯\frac{1}{3} \cdot \frac{1}{2}-\frac{1}{3} \cdot \frac{1}{2^2}+\frac{1}{3} \cdot \frac{1}{2^3}-\frac{1}{3} \cdot \frac{1}{2^4}+\cdots+(-1)^{n-1} \frac{1}{3} \cdot \frac{1}{2^n}+\cdots31⋅21−31⋅221+31⋅231−31⋅241+⋯+(−1)n−131⋅2n1+⋯;

(4) 1ln⁡2−1ln⁡3+1ln⁡4−1ln⁡5+⋯+(−1)n−11ln⁡(n+1)+⋯\frac{1}{\ln 2}-\frac{1}{\ln 3}+\frac{1}{\ln 4}-\frac{1}{\ln 5}+\cdots+(-1)^{n-1} \frac{1}{\ln (n+1)}+\cdotsln21−ln31+ln41−ln51+⋯+(−1)n−1ln(n+1)1+⋯;

(5) ∑n=1∞(−1)n+12n2n!\sum_{n=1}^{\infty}(-1)^{n+1} \frac{2^{n^2}}{n !}∑n=1∞(−1)n+1n!2n2.