Section 2.14

Series with Nonnegative Terms

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Subsection 2.14.2Problems[55/55%]
Problem 2.14.1

Prove the convergence of the series ∑n=1∞an\sum_{n = 1}^{\infty } a_{n} by establishing that the sequence of its partial sums is bounded above:

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(1)

an=sin⁡42n(n+1)(n+2)a_{n} = \frac{\sin^{4} 2 n}{(n+1)(n+2)}

(2)

an=π−arctg⁡n(n+1)(n+2)(n+3)a_{n} = \frac{\pi -\operatorname {arctg} n}{(n+1)(n+2)(n+3)}

(3)

an=(1+ln⁡nn)qn,0<q<1a_{n} = \left(1+\frac{\ln n}{n}\right) q^{n}, 0 < q < 1;

(4)

an=n2n+5n3n+4a_{n} = \frac{n 2^{n}+5}{n 3^{n}+4}.

Problem 2.14.2

Using the comparison test, investigate the series ∑n=1∞an\sum_{n = 1}^{\infty } a_{n} for convergence.

?
(1)

an=5+3(−1)n+12na_{n} = \frac{5+3(-1)^{n+1}}{2^{n}};

(2)

an=arctg⁡nn2+1a_{n} = \frac{\operatorname {arctg} n}{n^{2}+1}

(3)

an=sin⁡23nnna_{n} = \frac{\sin^{2} 3 n}{n \sqrt{n}}

(4)

an=cos⁡(π/4n)2n5−15a_{n} = \frac{\cos (\pi / 4 n)}{\sqrt[5]{2 n^{5}-1}}

(5)

an=sin⁡3+(−1)nn2a_{n} = \sin \frac{3+(-1)^{n}}{n^{2}}

(6)

an=ln⁡n+sin⁡nn2+2ln⁡na_{n} = \frac{\ln n+\sin n}{n^{2}+2 \ln n}

(7)

an=n+2n2(4+3sin⁡(πn/3))a_{n} = \frac{n+2}{n^{2}(4+3 \sin (\pi n / 3))};

(8)

an=ln⁡(1+(3+(−1)narctg⁡2n)/n)ln⁡2n,n⩾2a_{n} = \frac{\ln \left(1+(3+(-1)^{n} \operatorname {arctg} 2 n) / n\right)}{\ln^{2} n}, n \geqslant 2.

Problem 2.14.3

Using the comparison test, investigate the series ∑n=1∞an\sum_{n = 1}^{\infty } a_{n} for convergence.

?
(1)

an=arcsin⁡((n−1)/(n+1))nln⁡(n+1);a_{n} = \frac{\arcsin ((n-1) /(n+1))}{n \sqrt{\ln (n+1)}} ;

(2)

an=arctg⁡(n2+2n)3n+n2a_{n} = \frac{\operatorname {arctg}\left(n^{2}+2 n\right)}{3^{n}+n^{2}};

(3)

an=cos⁡4(2n/(n+1))n2+4−n2+1a_{n} = \frac{\cos^{4}(2 n /(n+1))}{\sqrt{n^{2}+4}-\sqrt{n^{2}+1}};

(4)

an=n5(2+sin⁡n)2n+n2a_{n} = \frac{n^{5}(\sqrt{2}+\sin \sqrt{n})}{2^{n}+n^{2}};

(5)

an=ln⁡(1+ln⁡n)n4+3n2+14ln⁡3(n+2)a_{n} = \frac{\ln (1+\ln n)}{\sqrt[4]{n^{4}+3 n^{2}+1} \ln^{3}(n+2)};

(6)

an=n2e−na_{n} = n^{2} e^{-n};

(7)

an=(3n+n3)e−nln⁡na_{n} = \left(3 n+n^{3}\right) e^{-\sqrt{n}} \ln n;

(8)

an=(3−2cos⁡2(πn/3))enn22na_{n} = \frac{\left(3-2 \cos^{2}(\pi n / 3)\right) e^{n}}{n^{2} 2^{n}}.

Problem 2.14.4

Investigate the convergence of the series ∑n=1∞an\sum_{n = 1}^{\infty } a_{n}, obtaining an asymptotic formula of the form an∼c/nαa_{n} \sim c / n^{\alpha } as n→∞n \rightarrow \infty.

?
(1)

an=1(2n+1)(2n+3)a_{n} = \frac{1}{\sqrt{(2 n+1)(2 n+3)}};

(2)

an=1−cos⁡2πna_{n} = 1-\cos \frac{2 \pi }{n};

(3)

an=ln⁡(1+1nn3)a_{n} = \ln \left(1+\frac{1}{n \sqrt[3]{n}}\right);

(4)

an=3n+1(2n+1)2a_{n} = \frac{3 n+1}{(2 n+1)^{2}};

(5)

an=1narctg⁡12na_{n} = \frac{1}{\sqrt{n}} \operatorname {arctg} \frac{1}{2 \sqrt{n}};

(6)

an=1n3arcsin⁡1n45a_{n} = \frac{1}{\sqrt[3]{n}} \arcsin \frac{1}{\sqrt[5]{n^{4}}};

(7)

an=(e1/n−1)sin⁡1n+1a_{n} = \left(e^{1 / n}-1\right) \sin \frac{1}{\sqrt{n+1}};

(8)

an=n3+3n2+5nn16+n4+116a_{n} = \frac{n^{3}+3 n^{2}+5}{n \sqrt[16]{n^{16}+n^{4}+1}}.

Problem 2.14.5

Investigate the convergence of the series ∑n=1∞an\sum_{n = 1}^{\infty } a_{n}, obtaining an asymptotic formula of the form an∼c/nαa_{n} \sim c / n^{\alpha } as n→∞n \rightarrow \infty.

?
(1)

an=sin⁡2n+1n3+5n+3a_{n} = \sin \frac{2 n+1}{n^{3}+5 n+3};

(2)

an=ntg⁡n+2n2+3a_{n} = n \operatorname {tg} \frac{n+2}{n^{2}+3}

(3)

an=exp⁡{n3+2n2+3}−1a_{n} = \exp \left\{ \frac{\sqrt[3]{n}+2}{n^{2}+3}\right\} -1;

(4)

an=ln⁡(1+sin⁡(1/n))n+ln⁡2na_{n} = \frac{\ln (1+\sin (1 / n))}{n+\ln^{2} n};

(5)

an=ln⁡n+3n2+4a_{n} = \ln \frac{n+3}{n^{2}+4}

(6)

an=ln⁡n2+4n2+3a_{n} = \ln \frac{n^{2}+4}{n^{2}+3}

(7)

an=n(ch⁡πn−1)a_{n} = \sqrt{n}\left(\operatorname {ch} \frac{\pi }{n}-1\right);

(8)

an=arcsin⁡(n+1)3n3+3n+2a_{n} = \arcsin \frac{(\sqrt{n}+1)^{3}}{n^{3}+3 n+2}.

Problem 2.14.6

Investigate the convergence of the series ∑n=1∞an\sum_{n = 1}^{\infty } a_{n}, obtaining an asymptotic formula of the form an∼c/nαa_{n} \sim c / n^{\alpha } as n→∞n \rightarrow \infty.

?
(1)

an=ln⁡1cos⁡(2π/n)a_{n} = \ln \frac{1}{\cos (2 \pi / n)};

(2)

an=ln⁡(1+2n)ln⁡n−1n+1,n⩾2a_{n} = \sqrt{\ln \left(1+\frac{2}{n}\right)} \ln \frac{\sqrt{n}-1}{\sqrt{n}+1}, n \geqslant 2;

(3)

an=n−1n2+1arctg⁡n+1n4+43a_{n} = \sqrt{\frac{n-1}{n^{2}+1}} \operatorname {arctg} \frac{n+1}{\sqrt[3]{n^{4}+4}};

(4)

an=nln⁡2n+12n−1−1a_{n} = n \ln \frac{2 n+1}{2 n-1}-1;

(5)

an=(1−ln⁡nn)2na_{n} = \left(1-\frac{\ln n}{n}\right)^{2 n};

(6)

an=log⁡2n(1+3nn)a_{n} = \log_{2^{n}}\left(1+\frac{\sqrt[n]{3}}{n}\right);

(7)

an=n+1−n−1n+2a_{n} = \frac{\sqrt{n+1}-\sqrt{n-1}}{\sqrt{n+2}};

(8)

an=n+1ln⁡ch⁡(1n)a_{n} = \sqrt{n+1} \ln \operatorname {ch}\left(\frac{1}{n}\right).

Problem 2.14.7

Find all values of α\alpha for which the series ∑n=1∞an\sum_{n = 1}^{\infty } a_{n} converges.

?
(1)

an=(1−nsin⁡(1/n))αa_{n} = (1-n \sin (1 / n))^{\alpha };

(2)

an=(sh⁡(1/n)−sin⁡(1/n))αa_{n} = (\operatorname {sh}(1 / n)-\sin (1 / n))^{\alpha };

(3)

an=(nsh⁡(1/n)−ch⁡(1/n))αa_{n} = (n \operatorname {sh}(1 / n)-\operatorname {ch}(1 / n))^{\alpha };

(4)

an=(−cos⁡(1/n)+ch⁡(1/n))αa_{n} = (-\cos (1 / n)+\operatorname {ch}(1 / n))^{\alpha };

(5)

an=(arctg⁡(1/n)−ln⁡(1+1/n))αa_{n} = (\operatorname {arctg}(1 / n)-\ln (1+1 / n))^{\alpha };

(6)

an=(etg⁡(1/n)−1)αa_{n} = \left(e^{\operatorname {tg}(1 / n)}-1\right)^{\alpha };

(7)

an=(exp⁡{1ncos⁡1n}−1−1n)αa_{n} = \left(\exp \left\{ \frac{1}{n} \cos \frac{1}{n}\right\} -1-\frac{1}{n}\right)^{\alpha };

(8)

an=(cos⁡1n−n2−nn)αa_{n} = \left(\cos \frac{1}{\sqrt{n}}-\frac{\sqrt{n^{2}-n}}{n}\right)^{\alpha }.

Problem 2.14.8

Find all values of α\alpha for which the series ∑n=1∞an\sum_{n = 1}^{\infty } a_{n} converges.

?
(1)

an=nsin⁡α(1n−arctg⁡1n)a_{n} = n \sin^{\alpha }\left(\frac{1}{n}-\operatorname {arctg} \frac{1}{n}\right);

(2)

an=(exp⁡{−12n2}−cos⁡1n)αa_{n} = \left(\exp \left\{ \frac{-1}{2 n^{2}}\right\} -\cos \frac{1}{n}\right)^{\alpha };

(3)

an=(ln⁡sh⁡1n−ln⁡1n)αa_{n} = \left(\ln \operatorname {sh} \frac{1}{n}-\ln \frac{1}{n}\right)^{\alpha };

(4)

an=∣ln⁡n+ln⁡sin⁡1n∣αa_{n} = \left|\ln n+\ln \sin \frac{1}{n}\right|^{\alpha };

(5)

an=(exp⁡{1−cos⁡1n}−1)αsin⁡1na_{n} = \left(\exp \left\{ 1-\cos \frac{1}{n}\right\} -1\right)^{\alpha } \sin \frac{1}{\sqrt{n}};

(6)

an=∣ln⁡n+1n−1−2n−1∣α,n⩾2a_{n} = \left|\ln \frac{n+1}{n-1}-\frac{2}{n-1}\right|^{\alpha }, n \geqslant 2;

(7)

an=ln⁡α(1+arctg⁡(1/n))sin⁡(n+1−n)a_{n} = \frac{\ln^{\alpha }(1+\sqrt{\operatorname {arctg}(1 / n)})}{\sin (\sqrt{n+1}-\sqrt{n})};

(8)

an=(n+1)(etg⁡(n+1−n)−1)αa_{n} = (n+1)\left(e^{\operatorname {tg}(\sqrt{n+1}-\sqrt{n})}-1\right)^{\alpha };

(9)

an=(nsin⁡1n−cos⁡1n3)αa_{n} = \left(n \sin \frac{1}{n}-\cos \frac{1}{n \sqrt{3}}\right)^{\alpha };

(10)

an=nα(ln⁡(n2+1)−2ln⁡n)a_{n} = n^{\alpha }\left(\ln \left(n^{2}+1\right)-2 \ln n\right);

(11)

an=n2+13−n2−13nαa_{n} = \frac{\sqrt[3]{n^{2}+1}-\sqrt[3]{n^{2}-1}}{n^{\alpha }};

(12)

an=(cos⁡cos⁡1n−cos⁡ch⁡1n)αa_{n} = \left(\cos \cos \frac{1}{n}-\cos \operatorname {ch} \frac{1}{n}\right)^{\alpha }.

Problem 2.14.9

Find all values of α\alpha for which the series ∑n=1∞an\sum_{n = 1}^{\infty } a_{n} converges.

?
(1)

an=(e1/2n−(1+sh⁡1n)1/2)αa_{n} = \left(e^{1 / 2 n}-\left(1+\operatorname {sh} \frac{1}{n}\right)^{1 / 2}\right)^{\alpha };

(2)

an=(1nsin⁡(1/n)−cos⁡1n)αa_{n} = \left(\frac{1}{n \sin (1 / n)}-\cos \frac{1}{n}\right)^{\alpha };

(3)

an=∣ln⁡arctg⁡1n−ln⁡tg⁡1n∣αa_{n} = \left|\ln \operatorname {arctg} \frac{1}{n}-\ln \operatorname {tg} \frac{1}{n}\right|^{\alpha };

(4)

an=(1−sin⁡πn22n2+1)αa_{n} = \left(1-\sin \frac{\pi n^{2}}{2 n^{2}+1}\right)^{\alpha };

(5)

an=(e2−(1+2/n)nn)αa_{n} = \left(\frac{e^{2}-(1+2 / n)^{n}}{n}\right)^{\alpha };

(6)

an=((sh⁡(1/n)sin⁡(1/n))3n−1)αa_{n} = \left(\left(\frac{\operatorname {sh}(1 / n)}{\sin (1 / n)}\right)^{3 n}-1\right)^{\alpha };

(7)

an=(n+1−n)αln⁡2n+12n−1a_{n} = (\sqrt{n+1}-\sqrt{n})^{\alpha } \ln \frac{2 n+1}{2 n-1};

(8)

an=(1−(cos⁡1n)1/n)αa_{n} = \left(1-\left(\cos \frac{1}{n}\right)^{1 / n}\right)^{\alpha }.

Problem 2.14.10

Find all values of α\alpha for which the series ∑n=1∞an\sum_{n = 1}^{\infty } a_{n} converges.

?
(1)

an=(n2+1)ln⁡αsh⁡(1/n)sin⁡(1/n)a_{n} = \left(n^{2}+1\right) \ln^{\alpha } \frac{\operatorname {sh}(1 / n)}{\sin (1 / n)};

(2)

an=(e−(1+1/n)n)α(1−cos⁡(1/n))2a_{n} = \frac{\left(e-(1+1 / n)^{n}\right)^{\alpha }}{(1-\cos (1 / n))^{2}};

(3)

an=(n2+n+14−n+12)αa_{n} = \left(\sqrt[4]{n^{2}+n+1}-\sqrt{n+\frac{1}{2}}\right)^{\alpha };

(4)

an=(1n−ln⁡n+1n)αa_{n} = \left(\frac{1}{\sqrt{n}}-\sqrt{\ln \frac{n+1}{n}}\right)^{\alpha };

(5)

an=ln⁡αch⁡(1/n)cos⁡(1/n);a_{n} = \ln^{\alpha } \frac{\operatorname {ch}(1 / n)}{\cos (1 / n)} ;

(6)

an=(n+n+n−nn)αa_{n} = \left(\frac{\sqrt{n+\sqrt{n+\sqrt{n}}}-\sqrt{n}}{n}\right)^{\alpha };

(7)

an=∣ln⁡cos⁡(sin⁡1n−arcsin⁡1n)∣αa_{n} = \left|\ln \cos \left(\sin \frac{1}{n}-\arcsin \frac{1}{n}\right)\right|^{\alpha };

(8)

an=(π2n+2−cos⁡πn2n+2)αa_{n} = \left(\frac{\pi }{2 n+2}-\cos \frac{\pi n}{2 n+2}\right)^{\alpha }.

Problem 2.14.11

Find all values of α\alpha for which the series ∑n=1∞an\sum_{n = 1}^{\infty } a_{n} converges.

?
(1)

an=α−ln⁡n,α>0a_{n} = \alpha^{-\ln n}, \alpha > 0;

(2)

an=(1+ln⁡nn)αna_{n} = \left(1+\frac{\ln n}{n}\right)^{\alpha n};

(3)

an=nα−(exp⁡{narcsin⁡1n2}−1)a_{n} = n^{\alpha }-\left(\exp \left\{ n \arcsin \frac{1}{n^{2}}\right\} -1\right);

(4)

an=exp⁡{narctg⁡1n2}−1−nαa_{n} = \exp \left\{ n \operatorname {arctg} \frac{1}{n^{2}}\right\} -1-n^{\alpha };

(5)

an=exp⁡{nsin⁡1n2}−nαa_{n} = \exp \left\{ n \sin \frac{1}{n^{2}}\right\} -n^{\alpha };

(6)

an=1n−ln⁡α(1+sin⁡1n)a_{n} = \frac{1}{n}-\ln^{\alpha }\left(1+\sin \frac{1}{n}\right);

(7)

an=(narcsin⁡1n)nα−1a_{n} = \left(n \arcsin \frac{1}{n}\right)^{n^{\alpha }}-1;

(8)

an=1−(nln⁡cos⁡1n+1)nαa_{n} = 1-\left(n \ln \cos \frac{1}{n}+1\right)^{n^{\alpha }}.

Problem 2.14.12

Find all values of α\alpha for which the series ∑n=1∞an\sum_{n = 1}^{\infty } a_{n} converges.

?
(1)

an=n2n(n+α)n+2(n+2)n+α,α>−1a_{n} = \frac{n^{2 n}}{(n+\alpha )^{n+2}(n+2)^{n+\alpha }}, \alpha > -1;

(2)

an=nln⁡(1+1nα)+arctg⁡12nα−1a_{n} = n \ln \left(1+\frac{1}{n^{\alpha }}\right)+\operatorname {arctg} \frac{1}{2 n^{\alpha }}-1;

(3)

an=exp⁡{12nα}−(1+sin⁡2(1nα/2))1/2a_{n} = \exp \left\{ \frac{1}{2 n^{\alpha }}\right\} -\left(1+\sin^{2}\left(\frac{1}{n^{\alpha / 2}}\right)\right)^{1 / 2};

(4)

an=(e1/n−sin⁡1n)nα−1a_{n} = \left(e^{1 / n}-\sin \frac{1}{n}\right)^{n^{\alpha }}-1;

(5)

an=ln⁡(1+tg⁡1n2α)−ln⁡cos⁡1nα,α>0a_{n} = \ln \left(1+\operatorname {tg} \frac{1}{n^{2 \alpha }}\right)-\ln \cos \frac{1}{n^{\alpha }}, \alpha > 0;

(6)

an=(ch⁡(1/n)cos⁡(1/n))nα−1a_{n} = \left(\frac{\operatorname {ch}(1 / n)}{\cos (1 / n)}\right)^{n^{\alpha }}-1;

(7)

an=ln⁡(1+ln⁡ch⁡1nα)−12n2α,α>0a_{n} = \ln \left(1+\ln \operatorname {ch} \frac{1}{n^{\alpha }}\right)-\frac{1}{2 n^{2 \alpha }}, \alpha > 0;

(8)

an=n+arctg⁡nα−na_{n} = \sqrt{n+\operatorname {arctg} n^{\alpha }}-\sqrt{n}.

Problem 2.14.13

Find all values of α\alpha for which the series ∑n=1∞an\sum_{n = 1}^{\infty } a_{n} converges.

?
(1)

an=1−(ln⁡(1−1n)+e1/n)nαa_{n} = 1-\left(\ln \left(1-\frac{1}{n}\right)+e^{1 / n}\right)^{n^{\alpha }};

(2)

an=n2n+1ln⁡(cos⁡1n+tg⁡2αn)a_{n} = \frac{n^{2}}{n+1} \ln \left(\cos \frac{1}{n}+\operatorname {tg}^{2} \frac{\alpha }{n}\right);

(3)

an=∣ln⁡(2π(arccos⁡1n+sin⁡1n))∣αa_{n} = \left|\ln \left(\frac{2}{\pi }\left(\arccos \frac{1}{n}+\sin \frac{1}{n}\right)\right)\right|^{\alpha };

(4)

an=1n(exp⁡{1nα+1}−1)a_{n} = \frac{1}{n}\left(\exp \left\{ \frac{1}{n^{\alpha }+1}\right\} -1\right);

(5)

an=(ln⁡n2+n+1n2−n+1−4n2+n/2+n2−n/2)αa_{n} = \left(\ln \frac{n^{2}+n+1}{n^{2}-n+1}-\frac{4}{\sqrt{n^{2}+n / 2}+\sqrt{n^{2}-n / 2}}\right)^{\alpha };

(6)

an=α1/n+α−1/n−2,α>0a_{n} = \alpha^{1 / n}+\alpha^{-1 / n}-2, \alpha > 0;

(7)

an=α1/n−α1/(n+1),α>0a_{n} = \alpha^{1 / n}-\alpha^{1 /(n+1)}, \alpha > 0.

Problem 2.14.14

Investigate the series ∑n=1∞an\sum_{n = 1}^{\infty } a_{n} for convergence, obtaining an asymptotic formula of the form am∼cnαln⁡βna_{m} \sim \frac{c}{n^{\alpha } \ln^{\beta } n} as n→∞n \rightarrow \infty:

?
(1)

an=nln⁡(n2+1)n5+3n+2a_{n} = \frac{n \ln \left(n^{2}+1\right)}{\sqrt{n^{5}+3 n+2}};

(2)

an=n+1(2n+1)2ln⁡2(n+1)a_{n} = \frac{\sqrt{n+1}}{(2 n+1)^{2} \ln^{2}(n+1)};

(3)

an=nn−1a_{n} = \sqrt[n]{n}-1;

(4)

an=cos⁡(1/n)nln⁡2tg⁡(1/n)a_{n} = \frac{\cos (1 / n)}{n \ln^{2} \operatorname {tg}(1 / n)};

(5)

an=(n−1n+1)nln⁡nn3+23a_{n} = \left(\frac{n-1}{n+1}\right)^{n} \frac{\ln n}{\sqrt[3]{n^{3}+2}};

(6)

an=ln⁡(en+n2)n2ln⁡2(n+1)a_{n} = \frac{\ln \left(e^{n}+n^{2}\right)}{n^{2} \ln^{2}(n+1)};

(7)

an=1−1ln⁡(n+1)na_{n} = 1-\frac{1}{\sqrt[n]{\ln (n+1)}};

(8)

an=1+ln⁡(2/n−ln⁡(1+1/n)2)ln⁡(n2+n)a_{n} = 1+\frac{\ln \left(2 / n-\ln (1+1 / n)^{2}\right)}{\ln \left(n^{2}+n\right)}.

Problem 2.14.15

Find all values of α\alpha for which the series ∑n=1∞an\sum_{n = 1}^{\infty } a_{n} converges:

?
(1)

an=ln⁡αch⁡(1/n)ln⁡2(n+1)a_{n} = \frac{\ln^{\alpha } \operatorname {ch}(1 / n)}{\ln^{2}(n+1)};

(2)

an=1nln⁡2(1+nα)a_{n} = \frac{1}{n \ln^{2}\left(1+n^{\alpha }\right)};

(3)

an=1n∣ln⁡(n2+1−n)∣αa_{n} = \frac{1}{n}\left|\ln \left(\sqrt{n^{2}+1}-n\right)\right|^{\alpha };

(4)

an=(arcsin⁡tg⁡(1/n))αln⁡(1+n+n2)a_{n} = \frac{(\arcsin \operatorname {tg}(1 / n))^{\alpha }}{\ln \left(1+n+n^{2}\right)};

(5)

an=arctg⁡(e−(1+1/n)n)ln⁡α(1+n2)a_{n} = \frac{\operatorname {arctg}\left(e-(1+1 / n)^{n}\right)}{\ln^{\alpha }\left(1+n^{2}\right)};

(6)

an=ln⁡(n2−n+1)(arcsin⁡(e1/n−1))αa_{n} = \frac{\ln \left(n^{2}-n+1\right)}{\left(\arcsin \left(e^{1 / \sqrt{n}}-1\right)\right)^{\alpha }};

(7)

an=ln⁡(n2+n+2)(n2+1−n)αa_{n} = \frac{\ln \left(n^{2}+n+2\right)}{\left(\sqrt{n^{2}+1}-n\right)^{\alpha }};

(8)

an=arcsin⁡((n+1)/(2n))−π/6ln⁡α(n+1)a_{n} = \frac{\arcsin ((n+1) /(2 n))-\pi / 6}{\ln^{\alpha }(n+1)};

(9)

an=(n1/(n2+1)−1)α;a_{n} = \left(n^{1 /\left(n^{2}+1\right)}-1\right)^{\alpha } ;

(10)

an=nnα−1a_{n} = n^{n^{\alpha }}-1;

(11)

an=n(ch⁡πn−cos⁡πnn2+1)αa_{n} = n\left(\operatorname {ch} \frac{\pi }{n}-\cos \frac{\pi n}{n^{2}+1}\right)^{\alpha };

(12)

an=((n+1)n−nn)αe−na_{n} = \left((n+1)^{n}-n^{n}\right)^{\alpha } e^{-n};

(13)

an=(n+narctg⁡n)αln⁡1+n22+n2a_{n} = (n+\sqrt{n \operatorname {arctg} n})^{\alpha } \ln \frac{1+n^{2}}{2+n^{2}};

(14)

an=sin⁡(1/(nα+n1−α))ln⁡2(1+n2)a_{n} = \frac{\sin \left(1 /\left(n^{\alpha }+n^{1-\alpha }\right)\right)}{\ln^{2}\left(1+n^{2}\right)};

(15)

an=(nn−1)αln⁡n,n>1a_{n} = \frac{(\sqrt[n]{n}-1)^{\alpha }}{\sqrt{\ln n}}, \quad n > 1;

(16)

an=(n−ctg⁡(1/n))αln⁡n3,n>1a_{n} = \frac{(n-\operatorname {ctg}(1 / n))^{\alpha }}{\sqrt[3]{\ln n}}, \quad n > 1;

(17)

an=(1−cos⁡(1/n)n)αln⁡3n,n>1a_{n} = \frac{(1-\sqrt[n]{\cos (1 / n)})^{\alpha }}{\ln^{3} n}, n > 1;

(18)

an=(e−(1+1/n2)n2)αln⁡2n,n>1a_{n} = \frac{\left(e-\left(1+1 / n^{2}\right)^{n^{2}}\right)^{\alpha }}{\ln^{2} n}, n > 1.

Problem 2.14.16

Having proved the convergence of the series ∑n=1∞(1n−ln⁡n+1n)\sum_{n = 1}^{\infty }\left(\frac{1}{n}-\ln \frac{n+1}{n}\right), obtain the asymptotic formula ∑k=1n1k=ln⁡n+C+αn\sum_{k = 1}^{n} \frac{1}{k} = \ln n+C+\alpha_{n}, where CC is Euler's constant (C=0,577215…),αn→0(C = 0,577215 \ldots ), \alpha_{n} \rightarrow 0 as n→∞n \rightarrow \infty.

?
Problem 2.14.17

Determine what condition the positive numbers a,b,ca, b, c must satisfy for the series ∑n=1∞an\sum_{n = 1}^{\infty } a_{n} to converge, where

an=2a1/n−b1/n−c1/n a_{n} = 2 a^{1 / n}-b^{1 / n}-c^{1 / n}
?
Problem 2.14.18

Investigate the series ∑n=1∞an\sum_{n = 1}^{\infty } a_{n} for convergence using the d'Alembert test.

?
(1)

an=n10(n+1)!;a_{n} = \frac{n^{10}}{(n+1) !} ;

(2)

an=n33na_{n} = \frac{n^{3}}{3^{n}};

(3)

an=2⋅5⋅8…(3n−1)1⋅6⋅11…(5n−4)a_{n} = \frac{2 \cdot 5 \cdot 8 \ldots (3 n-1)}{1 \cdot 6 \cdot 11 \ldots (5 n-4)};

(4)

an=n!annn,a≠e,a>0a_{n} = \frac{n ! a^{n}}{n^{n}}, a \neq e, \quad a > 0;

(5)

an=1⋅5…(4n−3)2⋅6…(4n−2)a_{n} = \frac{1 \cdot 5 \ldots (4 n-3)}{2 \cdot 6 \ldots (4 n-2)};

(6)

an=1⋅3⋅5…(2n−1)3nn!a_{n} = \frac{1 \cdot 3 \cdot 5 \ldots (2 n-1)}{3^{n} n !};

(7)

an=4⋅7⋅10…(3n+4)2⋅6⋅10…(4n+2)a_{n} = \frac{4 \cdot 7 \cdot 10 \ldots (3 n+4)}{2 \cdot 6 \cdot 10 \ldots (4 n+2)};

(8)

an=(2n)!(n!)2a_{n} = \frac{(2 n) !}{(n !)^{2}}.

Problem 2.14.19

Investigate the series ∑n=1∞an\sum_{n = 1}^{\infty } a_{n} for convergence using the d'Alembert test.

?
(1)

an=nnn!(2,7)n+1a_{n} = \frac{n^{n}}{n !(2,7)^{n+1}};

(2)

an=(2n+1)!(3n+4)3na_{n} = \frac{(2 n+1) !}{(3 n+4) 3^{n}};

(3)

an=(2n+1)!!1⋅…(3n+1)a_{n} = \frac{(2 n+1) ! !}{1 \cdot \ldots (3 n+1)};

(4)

an=a(a+1)…(a+(n−1))(2n−1)!!a_{n} = \frac{a(a+1) \ldots (a+(n-1))}{(2 n-1) ! !}, a>0a > 0;

(5)

an=2⋅5…(3n+2)2n(n+1)!a_{n} = \frac{2 \cdot 5 \ldots (3 n+2)}{2^{n}(n+1) !};

(6)

an=(2n+1)!!3nn!a_{n} = \frac{(2 n+1) ! !}{3^{n} n !};

(7)

an=3⋅6…(3n)(n+1)!arcsin⁡12na_{n} = \frac{3 \cdot 6 \ldots (3 n)}{(n+1) !} \arcsin \frac{1}{2^{n}};

(8)

an=(2n)!!n!arctg⁡13na_{n} = \frac{(2 n) ! !}{n !} \operatorname {arctg} \frac{1}{3^{n}};

(9)

an=52n(n!)3(3n)!;a_{n} = \frac{5^{2 n}(n !)^{3}}{(3 n) !} ;

(10)

an=(2n)!n!(n+1)!32na_{n} = \frac{(2 n) !}{n !(n+1) ! 3^{2 n}};

(11)

an=n2n(2n)!52n(n!)4a_{n} = \frac{n^{2 n}(2 n) !}{5^{2 n}(n !)^{4}};

(12)

an=(5n)!25n(2n)!(3n)!a_{n} = \frac{(5 n) !}{25^{n}(2 n) !(3 n) !};

(13)

an=32n(n!)4(3n)!(n+1)!a_{n} = \frac{3^{2 n}(n !)^{4}}{(3 n) !(n+1) !};

(14)

an=32n(2n)!nnn!a_{n} = \frac{3^{2 n}(2 n) !}{n^{n} n !};

(15)

an=n!ch⁡n(2n)!a_{n} = \frac{n ! \operatorname {ch} n}{\sqrt{(2 n) !}}

(16)

an=n2n(2n)!sh⁡na_{n} = \frac{n^{2 n}}{(2 n) ! \operatorname {sh} n}.

Problem 2.14.20

Investigate the series ∑n=1∞an\sum_{n = 1}^{\infty } a_{n} for convergence using the d'Alembert test.

?
(1)

an=(3n)!(n!)343na_{n} = \frac{(3 n) !}{(n !)^{3} 4^{3 n}};

(2)

an=n!(2n+1)!(3n)!a_{n} = \frac{n !(2 n+1) !}{(3 n) !}.

Problem 2.14.21

Investigate the series ∑n=1∞an\sum_{n = 1}^{\infty } a_{n} for convergence using the Cauchy test.

?
(1)

an=1(ln⁡n)n,n⩾2a_{n} = \frac{1}{(\ln n)^{n}}, \quad n \geqslant 2;

(2)

an=(3n)na_{n} = \left(\frac{3}{n}\right)^{n}

(3)

an=(ann+2)n,a>0a_{n} = \left(\frac{a n}{n+2}\right)^{n}, \quad a > 0;

(4)

an=2n(nn+1)n2a_{n} = 2^{n}\left(\frac{n}{n+1}\right)^{n^{2}}

(5)

an=(n+2n+3)n3/2a_{n} = \left(\frac{\sqrt{n}+2}{\sqrt{n}+3}\right)^{n^{3 / 2}};

(6)

an=3n+1(n+2n+3)n2a_{n} = 3^{n+1}\left(\frac{n+2}{n+3}\right)^{n^{2}}

(7)

an=(n2+5n2+6)n3a_{n} = \left(\frac{n^{2}+5}{n^{2}+6}\right)^{n^{3}};

(8)

an=(n−1n+1)n2+4n+5a_{n} = \left(\frac{n-1}{n+1}\right)^{n^{2}+4 n+5}

(9)

an=(cos⁡1n)n2a_{n} = \left(\cos \frac{1}{\sqrt{n}}\right)^{n^{2}};

(10)

an=(nsin⁡1n)n3a_{n} = \left(n \sin \frac{1}{n}\right)^{n^{3}}

(11)

an=(1ch⁡(1/n))n2a_{n} = \left(\frac{1}{\operatorname {ch}(1 / \sqrt{n})}\right)^{n^{2}};

(12)

an=(nsh⁡1n)−n3a_{n} = \left(n \operatorname {sh} \frac{1}{n}\right)^{-n^{3}};

(13)

an=13n(n+2n)n2a_{n} = \frac{1}{3^{n}}\left(\frac{n+2}{n}\right)^{n^{2}};

(14)

an=(narcsin⁡1n)n3a_{n} = \left(n \arcsin \frac{1}{n}\right)^{n^{3}}.

Problem 2.14.22

Investigate the series ∑n=1∞an\sum_{n = 1}^{\infty } a_{n} for convergence using the Cauchy test.

?
(1)

an=3−n(n+1n)n2a_{n} = 3^{-n}\left(\frac{n+1}{n}\right)^{n^{2}};

(2)

an=(2n−12n+1)n(n−1)a_{n} = \left(\frac{2 n-1}{2 n+1}\right)^{n(n-1)};

(3)

an=nn+1(3n2+2n+1)(n+3)/2a_{n} = \frac{n^{n+1}}{\left(3 n^{2}+2 n+1\right)^{(n+3) / 2}};

(4)

an=(n−1n+1)n3+3n+1a_{n} = \left(\frac{n-1}{n+1}\right)^{\sqrt{n^{3}+3 n+1}}

(5)

an=nα(ln⁡(n+1))n/2,α>0a_{n} = \frac{n^{\alpha }}{(\ln (n+1))^{n / 2}}, \quad \alpha > 0;

(6)

an=(6n+15n−3)n/2(56)2n/3a_{n} = \left(\frac{6 n+1}{5 n-3}\right)^{n / 2}\left(\frac{5}{6}\right)^{2 n / 3}.

Problem 2.14.23

Investigate the series ∑n=1∞an\sum_{n = 1}^{\infty } a_{n} for convergence using the Raabe test or the Gauss test:

?
(1)

an=(2n−1)!!(2n)!!a_{n} = \frac{(2 n-1) ! !}{(2 n) ! !};

(2)

an=((2n+1)!!(2n+2)!!)α1nβa_{n} = \left(\frac{(2 n+1) ! !}{(2 n+2) ! !}\right)^{\alpha } \frac{1}{n^{\beta }};

(3)

an=((a+1)(a+2)…(a+n)(b+1)(b+2)…(b+n))α,a>0,b>0a_{n} = \left(\frac{(a+1)(a+2) \ldots (a+n)}{(b+1)(b+2) \ldots (b+n)}\right)^{\alpha }, a > 0, b > 0;

(4)

an=n!ennn+αa_{n} = \frac{n ! e^{n}}{n^{n+\alpha }}

(5)

an=n!(a+2)(a+3)…(a+n+1),a>0a_{n} = \frac{\sqrt{n !}}{(a+\sqrt{2})(a+\sqrt{3}) \ldots (a+\sqrt{n+1})}, \quad a > 0;

(6)

an=(n+1)!β(β+1)…(β+n)nα,β>0a_{n} = \frac{(n+1) !}{\beta (\beta +1) \ldots (\beta +n) n^{\alpha }}, \quad \beta > 0;

(7)

an=1⋅4…(3n−2)⋅2⋅5…(3n+2)n!(n+1)!9na_{n} = \frac{1 \cdot 4 \ldots (3 n-2) \cdot 2 \cdot 5 \ldots (3 n+2)}{n !(n+1) ! 9^{n}};

(8)

an=ln⁡2ln⁡3…ln⁡(n+1)ln⁡(2+a)ln⁡(3+a)…ln⁡(n+1+a),a>0a_{n} = \frac{\ln 2 \ln 3 \ldots \ln (n+1)}{\ln (2+a) \ln (3+a) \ldots \ln (n+1+a)}, \quad a > 0.

Problem 2.14.24

Investigate the series ∑n=1∞an\sum_{n = 1}^{\infty } a_{n} for convergence.

?
(1)

an=1nsin⁡πna_{n} = \frac{1}{\sqrt{n}} \sin \frac{\pi }{n}

(2)

an=1n3ln⁡(1+1n+1)a_{n} = \frac{1}{\sqrt[3]{n}} \ln \left(1+\frac{1}{\sqrt{n+1}}\right);

(3)

an=nsin⁡22nn5+3a_{n} = \frac{n \sin^{2} 2 n}{\sqrt{n^{5}+3}}

(4)

an=arctg⁡n+2nln⁡2(n+1)a_{n} = \frac{\operatorname {arctg} \sqrt{n+2}}{n \ln^{2}(n+1)};

(5)

an=arctg⁡(3+(−1)n2)ntg⁡1na_{n} = \frac{\operatorname {arctg}\left(3+(-1)^{n} 2\right)}{\sqrt{n}} \operatorname {tg} \frac{1}{\sqrt{n}};

(6)

an=n2+3n3(3−2sin⁡(πn/3))a_{n} = \frac{n^{2}+3}{n^{3}(3-2 \sin (\pi n / 3))};

(7)

an=arcsin⁡((n+1)/2n)3n4+24a_{n} = \frac{\arcsin ((n+1) / 2 n)}{\sqrt[4]{3 n^{4}+2}};

(8)

an=en/(n2+1)−1a_{n} = e^{\sqrt{n} /\left(n^{2}+1\right)}-1.

Problem 2.14.25

Investigate the series ∑n=1∞an\sum_{n = 1}^{\infty } a_{n} for convergence.

?
(1)

an=tg⁡1n−arctg⁡1na_{n} = \operatorname {tg} \frac{1}{n}-\operatorname {arctg} \frac{1}{n};

(2)

an=n+2n+3ln⁡3n−13n+1a_{n} = \frac{\sqrt{n+2}}{n+3} \ln \frac{3 n-1}{3 n+1};

(3)

an=(3n+12n+3)na_{n} = \left(\frac{3 n+1}{2 n+3}\right)^{n};

(4)

an=1n3(3n2+42n2+3)na_{n} = \frac{1}{n^{3}}\left(\frac{3 n^{2}+4}{2 n^{2}+3}\right)^{n};

(5)

an=n5e−na_{n} = n^{5} e^{-\sqrt{n}}

(6)

an=nαqn,0<q<1a_{n} = n^{\alpha } q^{n}, 0 < q < 1;

(7)

an=(n!)2(2n)!a_{n} = \frac{(n !)^{2}}{(2 n) !}

(8)

an=(n+1)!7n(n2+4)3a_{n} = \frac{(n+1) !}{7^{n}\left(n^{2}+4\right)^{3}};

(9)

an=1nαln⁡βn,n⩾2a_{n} = \frac{1}{n^{\alpha } \ln^{\beta } n}, \quad n \geqslant 2;

(10)

an=sin⁡n2(3+1/n)na_{n} = \sin \frac{n^{2}}{(3+1 / n)^{n}};

(11)

an=tg⁡(n+13n+2)na_{n} = \operatorname {tg}\left(\frac{n+1}{3 n+2}\right)^{n};

(12)

an=(n+1n)n213na_{n} = \left(\frac{n+1}{n}\right)^{n^{2}} \frac{1}{3^{n}};

(13)

an=(n2+n−n)n2a_{n} = \left(\sqrt{n^{2}+n}-n\right)^{n^{2}};

(14)

an=(n2+1−n+1)−n2a_{n} = \left(\sqrt{n^{2}+1}-n+1\right)^{-n^{2}};

(15)

an=2n(n−1n)n2a_{n} = 2^{n}\left(\frac{n-1}{n}\right)^{n^{2}};

(16)

an=(narctg⁡1n2)n5a_{n} = \left(n \operatorname {arctg} \frac{1}{n^{2}}\right)^{n^{5}}.

Problem 2.14.26

Investigate the series ∑n=1∞an\sum_{n = 1}^{\infty } a_{n} for convergence.

?
(1)

an=(2,6)nn!nna_{n} = \frac{(2,6)^{n} n !}{n^{n}};

(2)

an=n−n/3n!+13a_{n} = n^{-n / 3} \sqrt[3]{n !+1};

(3)

an=3ncos⁡1n!a_{n} = 3^{n} \cos \frac{1}{n !};

(4)

an=nn(n!)2a_{n} = \frac{n^{n}}{(n !)^{2}};

(5)

an=(2n+2)!πn(n!)2a_{n} = \frac{(2 n+2) !}{\pi^{n}(n !)^{2}}

(6)

an=92nn!(2n)!a_{n} = \frac{9^{2 n} n !}{(2 n) !}

(7)

an=α(α+1)…(α+1)n!nβ,α>0a_{n} = \frac{\alpha (\alpha +1) \ldots (\alpha +1)}{n ! n^{\beta }}, \alpha > 0;

(8)

an=a(a+c)(a+2c)…(a+nc)b(b+c)(b+2c)…(b+nc),a>0,b>0,c>0a_{n} = \frac{a(a+c)(a+2 c) \ldots (a+n c)}{b(b+c)(b+2 c) \ldots (b+n c)}, \quad a > 0, \quad b > 0, \quad c > 0;

(9)

an=nn2/2(arctg⁡1n)n2a_{n} = n^{n^{2} / 2}\left(\operatorname {arctg} \frac{1}{\sqrt{n}}\right)^{n^{2}};

(10)

(ntg⁡1n)nn2+n\left(\sqrt{n} \operatorname {tg} \frac{1}{\sqrt{n}}\right)^{n \sqrt{n^{2}+n}};

(11)

an=n!arctg⁡2nnna_{n} = n ! \operatorname {arctg} \frac{2^{n}}{n^{n}}.

Problem 2.14.27

Investigate the series ∑n=1∞an\sum_{n = 1}^{\infty } a_{n} for convergence.

?
(1)

an=1ln⁡(n+1)na_{n} = \frac{1}{\sqrt[n]{\ln (n+1)}};

(2)

an=(2−23)(2−25)…(2−22n+1)a_{n} = (\sqrt{2}-\sqrt[3]{2})(\sqrt{2}-\sqrt[5]{2}) \ldots (\sqrt{2}-\sqrt[2 n+1]{2});

(3)

an=(n!)33n4/3a_{n} = \frac{(n !)^{3}}{3^{n^{4 / 3}}};

(4)

an=1(ln⁡n)ln⁡n,n⩾2a_{n} = \frac{1}{(\ln n)^{\ln n}}, \quad n \geqslant 2;

(5)

an=1(ln⁡ln⁡n)ln⁡n,n⩾3a_{n} = \frac{1}{(\ln \ln n)^{\ln n}}, \quad n \geqslant 3

(6)

an=1(ln⁡n)ln⁡ln⁡n,n⩾3a_{n} = \frac{1}{(\ln n)^{\ln \ln n}}, \quad n \geqslant 3;

(7)

an=nln⁡n(ln⁡n)n,n⩾2a_{n} = \frac{n^{\ln n}}{(\ln n)^{n}}, \quad n \geqslant 2;

(8)

an=((n+1)!)n2!4!…(2n)!a_{n} = \frac{((n+1) !)^{n}}{2 ! 4 ! \ldots (2 n) !}

(9)

an=nn(2n)!ch⁡na_{n} = \frac{n^{n}}{\sqrt{(2 n) !} \operatorname {ch} n}

(10)

an=(arctg⁡1n3)n(3n)!a_{n} = \left(\operatorname {arctg} \frac{1}{n^{3}}\right)^{n}(3 n) !;

(11)

an=12⋅42⋅72…(3n+1)2(2n)!arctg⁡12n;a_{n} = \frac{1^{2} \cdot 4^{2} \cdot 7^{2} \ldots (3 n+1)^{2}}{(2 n) !} \operatorname {arctg} \frac{1}{2^{n}} ;

(12)

an=22⋅52⋅82…(3n+2)2(2n+1)!sin⁡13na_{n} = \frac{2^{2} \cdot 5^{2} \cdot 8^{2} \ldots (3 n+2)^{2}}{(2 n+1) !} \sin \frac{1}{3^{n}}.

Problem 2.14.28

Investigate the series ∑n=1∞an\sum_{n = 1}^{\infty } a_{n} for convergence.

?
(1)

an=1ln⁡(n!),n⩾2a_{n} = \frac{1}{\ln (n !)}, \quad n \geqslant 2;

(2)

an=(∑k=1nln⁡2k)/nαa_{n} = \left(\sum_{k = 1}^{n} \ln^{2} k\right) / n^{\alpha }

(3)

an=ln⁡(n!)nα,n⩾2a_{n} = \frac{\ln (n !)}{n^{\alpha }}, \quad n \geqslant 2;

(4)

an=1n(ln⁡n)α(ln⁡ln⁡n)β,n>2a_{n} = \frac{1}{n(\ln n)^{\alpha }(\ln \ln n)^{\beta }}, n > 2.

Problem 2.14.29

Having established the convergence of the corresponding series, show that:

?
(1)

lim⁡n→∞n!nn=0\lim_{n \rightarrow \infty } \frac{n !}{n^{n}} = 0;

(2)

lim⁡n→∞nn(n!)2=0\lim_{n \rightarrow \infty } \frac{n^{n}}{(n !)^{2}} = 0;

(3)

lim⁡n→∞(n!)nnn2=0\lim_{n \rightarrow \infty } \frac{(n !)^{n}}{n^{n^{2}}} = 0;

(4)

lim⁡n→∞(2n)!an=0,a>1\lim_{n \rightarrow \infty } \frac{(2 n) !}{a^{n}} = 0, a > 1.

Problem 2.14.30

Investigate the series ∑n=1∞μ(n)n2\sum_{n = 1}^{\infty } \frac{\mu (n)}{n^{2}} for convergence, where μ(n)\mu (n) is the number of the digits of the number nn.

?
Problem 2.14.31

Let a1=1,a2=1,an+1=an+1+an(n⩾1)a_{1} = 1, a_{2} = 1, a_{n+1} = a_{n+1}+a_{n}(n \geqslant 1). Prove that the series ∑n=1∞1an\sum_{n = 1}^{\infty } \frac{1}{a_{n}} converges.

?
Problem 2.14.32

Let λ1,λ2,…,λn,…\lambda_{1}, \lambda_{2}, \ldots , \lambda_{n}, \ldots be the positive roots of the equation tg⁡x=x\operatorname {tg} x = x, arranged in increasing order. Prove that the series ∑n=1∞λn−2\sum_{n = 1}^{\infty } \lambda_{n}^{-2} converges.

?
Problem 2.14.33

Investigate the series ∑n=1∞an\sum_{n = 1}^{\infty } a_{n} for convergence:

?
(1)

an=∫01/nx31+x4dxa_{n} = \int_{0}^{1 / n} \frac{\sqrt[3]{x}}{1+x^{4}} d x;

(2)

an=∫nn+2e−x4dx;a_{n} = \int_{n}^{n+2} e^{-\sqrt[4]{x}} d x ;

(3)

an=∫πnπ+πncos⁡2xxdxa_{n} = \int_{\pi n}^{\pi +\pi n} \frac{\cos^{2} x}{x} d x;

(4)

an=∫0π/nxsin⁡5x1+x2dxa_{n} = \int_{0}^{\pi / n} \frac{x \sin^{5} x}{1+x^{2}} d x

Problem 2.14.34

Let a1=2,a2=2−2,a3=2−2+2,…,an=2−2+2+…+2a_{1} = \sqrt{2}, a_{2} = \sqrt{2-\sqrt{2}}, a_{3} = \sqrt{2-\sqrt{2+\sqrt{2}}}, \ldots , a_{n} = \sqrt{2-\sqrt{2+\sqrt{2+\ldots +\sqrt{2}}}}. Prove that the series ∑n=1∞an\sum_{n = 1}^{\infty } a_{n} converges.

?
Problem 2.14.35

Prove that if an⩾0,bn>0a_{n} \geqslant 0, b_{n} > 0 for all n⩾n0n \geqslant n_{0}, the series ∑n=1∞bn\sum_{n = 1}^{\infty } b_{n} converges and lim⁡n→∞anbn=0\lim_{n \rightarrow \infty } \frac{a_{n}}{b_{n}} = 0, then the series ∑n=1∞an\sum_{n = 1}^{\infty } a_{n} also converges.

?
Problem 2.14.36

Prove that if an⩾0,bn>0a_{n} \geqslant 0, b_{n} > 0 for all n⩾n0n \geqslant n_{0}, the series ∑n=1∞bn\sum_{n = 1}^{\infty } b_{n} diverges, lim⁡n→∞anbn=∞\lim_{n \rightarrow \infty } \frac{a_{n}}{b_{n}} = \infty, then the series ∑n=1∞an\sum_{n = 1}^{\infty } a_{n} also diverges.

?
Problem 2.14.37

Prove that if lim⁡n→∞nan=a\lim_{n \rightarrow \infty } n a_{n} = a exists, where a≠0a \neq 0, then the series ∑n=1∞an\sum_{n = 1}^{\infty } a_{n} also diverges.

?
Problem 2.14.38

Prove that if an>0,an+1⩽ana_{n} > 0, a_{n+1} \leqslant a_{n} for all n∈Nn \in N and the series ∑n=1∞an\sum_{n = 1}^{\infty } a_{n} converges, then lim⁡n→∞nan=0\lim_{n \rightarrow \infty } n a_{n} = 0.

?
Problem 2.14.39

Prove that if an⩾0(n∈N)a_{n} \geqslant 0(n \in N) and the series ∑n=1∞an\sum_{n = 1}^{\infty } a_{n} converges, then the series ∑n=1∞an2\sum_{n = 1}^{\infty } a_{n}^{2} also converges. Is the converse statement true?

?
Problem 2.14.40

Prove that a series with nonnegative terms converges if at least one subsequence of the sequence of its partial sums is bounded above.

?
Problem 2.14.41

Prove that if the sequence {nan}\left\{ n a_{n}\right\}, where an⩾0  (n∈N)a_{n} \geqslant 0 \; (n \in N), is bounded, then the series ∑n=1∞an2\sum_{n = 1}^{\infty } a_{n}^{2} converges.

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Problem 2.14.42

Prove that if the series ∑n=1∞ann\sum_{n = 1}^{\infty } \frac{a_{n}}{\sqrt{n}}, where an⩾0(n∈N)a_{n} \geqslant 0(n \in N), converges and an+1⩽an(n∈N)a_{n+1} \leqslant a_{n}(n \in N), then the series ∑n=1∞an2\sum_{n = 1}^{\infty } a_{n}^{2} is also convergent.

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Problem 2.14.43

Prove that if an⩾0,bn⩾0,cn⩾0(n∈N)a_{n} \geqslant 0, b_{n} \geqslant 0, c_{n} \geqslant 0(n \in N) and the series ∑n=1∞an3,∑n=1∞bn3,∑n=1∞cn3\sum_{n = 1}^{\infty } a_{n}^{3}, \sum_{n = 1}^{\infty } b_{n}^{3}, \sum_{n = 1}^{\infty } c_{n}^{3} converge, then the series ∑n=1∞anbncn\sum_{n = 1}^{\infty } a_{n} b_{n} c_{n} also converges.

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Problem 2.14.44

Let an⩾0,bn⩾0(n∈N)a_{n} \geqslant 0, b_{n} \geqslant 0(n \in N) and the series ∑n=1∞an,∑n=1∞bn\sum_{n = 1}^{\infty } a_{n}, \sum_{n = 1}^{\infty } b_{n} diverge. Determine whether it follows that the following series diverges:

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(1)

∑n=1∞max⁡(an,bn)\sum_{n = 1}^{\infty } \max \left(a_{n}, b_{n}\right);

(2)

∑n=1∞min⁡(an,bn)\sum_{n = 1}^{\infty } \min \left(a_{n}, b_{n}\right).

Problem 2.14.45

Prove that if an>0(n∈N)a_{n} > 0(n \in N) and there exists

lim⁡n→∞an+1an=q \lim _{n \rightarrow \infty } \frac{a_{n+1}}{a_{n}} = q

then an=o(q1n)a_{n} = o\left(q_{1}^{n}\right), where q1>qq_{1} > q.

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Problem 2.14.46

Prove that if an>0(n∈N)a_{n} > 0(n \in N) and for all n⩾mn \geqslant m the inequality an+1/an⩽λ<1a_{n+1} / a_{n} \leqslant \lambda < 1 holds, then

∑k=n+1∞ak⩽amλn−m+11−λ \sum _{k = n+1}^{\infty } a_{k} \leqslant a_{m} \frac{\lambda ^{n-m+1}}{1-\lambda }

for all n⩾mn \geqslant m.

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Problem 2.14.47

Prove that if an>0(n∈N)a_{n} > 0(n \in N) and there exists

lim‾⁡n→∞an+1an=q<1, \varlimsup _{n \rightarrow \infty } \frac{a_{n+1}}{a_{n}} = q < 1,

then the series ∑n=1∞an\sum_{n = 1}^{\infty } a_{n} converges (generalized d'Alembert test).

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Problem 2.14.48

Prove that if an⩾0(n∈N)a_{n} \geqslant 0(n \in N) and there exists

lim‾⁡n→∞ann=q, \varlimsup _{n \rightarrow \infty } \sqrt[n]{a_{n}} = q,

then the series ∑n=1∞an\sum_{n = 1}^{\infty } a_{n} converges when q<1q < 1, and diverges when q>1q > 1 (generalized Cauchy test).

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Problem 2.14.49

Prove that if an>0,an+1⩽an(n∈N)a_{n} > 0, a_{n+1} \leqslant a_{n}(n \in N) and lim⁡n→∞a2nan=q\lim_{n \rightarrow \infty } \frac{a_{2 n}}{a_{n}} = q exists, then the series ∑n=1∞an\sum_{n = 1}^{\infty } a_{n} converges when q<12q < \frac{1}{2}, and diverges when q>12q > \frac{1}{2}.

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Problem 2.14.50

Prove that if an⩾0(n∈N)a_{n} \geqslant 0(n \in N) and there exists an index mm such that for all n>mn > m the inequality holds

nln⁡n(1−ann)⩾λ>1, \frac{n}{\ln n}\left(1-\sqrt[n]{a_{n}}\right) \geqslant \lambda > 1,

then the series ∑n=1∞an\sum_{n = 1}^{\infty } a_{n} converges; but if

nln⁡n(1−ann)⩽1 \frac{n}{\ln n}\left(1-\sqrt[n]{a_{n}}\right) \leqslant 1

for all n>mn > m, then the series ∑n=1∞an\sum_{n = 1}^{\infty } a_{n} diverges (Jamet's test).

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Problem 2.14.51

Prove that if an>0(n∈N)a_{n} > 0(n \in N) and there exist an index mm and a number α>0\alpha > 0 such that for all n⩾mn \geqslant m the inequality holds

ln⁡an−1ln⁡n⩾1+α, \frac{\ln a_{n}^{-1}}{\ln n} \geqslant 1+\alpha ,

then the series ∑n=1∞an\sum_{n = 1}^{\infty } a_{n} converges; but if

ln⁡an−1ln⁡n⩽1 \frac{\ln a_{n}^{-1}}{\ln n} \leqslant 1

for all n⩾mn \geqslant m, then the series ∑n=1∞an\sum_{n = 1}^{\infty } a_{n} diverges (logarithmic test).

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Problem 2.14.52

Prove that if an>0,an+1⩽an(n∈N)a_{n} > 0, a_{n+1} \leqslant a_{n}(n \in N), then the series ∑n=1∞an\sum_{n = 1}^{\infty } a_{n} converges or diverges together with the series ∑n=1∞2na2n\sum_{n = 1}^{\infty } 2^{n} a_{2^{n}}.

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Problem 2.14.53

Prove that the series ∑n=1∞f(n)\sum_{n = 1}^{\infty } f(n), where f(x)f(x) is a positive and decreasing function on the interval [1;+∞)[1 ;+\infty ) such that there exists

lim⁡x→+∞exf(ex)f(x)=λ \lim _{x \rightarrow +\infty } \frac{e^{x} f\left(e^{x}\right)}{f(x)} = \lambda

converges when λ<1\lambda < 1 and diverges when λ>1\lambda > 1 (Yermakov's test).

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Problem 2.14.54

Prove that if an>0,an+1⩽an(n∈N),lim⁡n→∞an=0a_{n} > 0, a_{n+1} \leqslant a_{n}(n \in N), \lim_{n \rightarrow \infty } a_{n} = 0, then the series ∑n=1∞an\sum_{n = 1}^{\infty } a_{n} converges or diverges together with the series ∑m=1∞qm2−m\sum_{m = 1}^{\infty } q_{m} 2^{-m}, where qmq_{m} is the largest index of the terms of the series ∑n=1∞an\sum_{n = 1}^{\infty } a_{n} satisfying the condition

an⩾2−m(n=1,2,…,qm) a_{n} \geqslant 2^{-m} \quad \left(n = 1,2, \ldots , q_{m}\right)

(Lobachevsky's test).

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Problem 2.14.55

Prove that if f(x)f(x) is a positive, decreasing function on the interval [1;+∞)[1 ;+\infty ) and the series ∑n=1∞fn\sum_{n = 1}^{\infty } f_{n} converges, then for its nn-th remainder

rn=∑k=n+1∞f(k) r_{n} = \sum _{k = n+1}^{\infty } f(k)

the inequalities ∫n+1+∞f(x)dx<rn<f(n+1)+∫n+1∞f(x)dx\int_{n+1}^{+\infty } f(x) d x < r_{n} < f(n+1)+\int_{n+1}^{\infty } f(x) d x hold.

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