Section 2.15

Absolutely and Non-Absolutely Convergent Series

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Subsection 2.15.2Problems[36/44%]
Problem 2.15.1

Prove that the series converge absolutely.

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(1)

∑n=1∞sin⁡(2n+π/4)nn+23\sum_{n = 1}^{\infty } \frac{\sin (2 n+\pi / 4)}{n \sqrt[3]{n+2}};

(2)

∑n=1∞arctg⁡(−n)n2n6+3n+14\sum_{n = 1}^{\infty } \frac{\operatorname {arctg}(-n)^{n}}{\sqrt[4]{2 n^{6}+3 n+1}}

(3)

∑n=1∞cos⁡(πn/4)(n+2)ln⁡3(n+3)\sum_{n = 1}^{\infty } \frac{\cos (\pi n / 4)}{(n+2) \sqrt{\ln^{3}(n+3)}};

(4)

∑n=1∞(−1)n+1ln⁡2n2n\sum_{n = 1}^{\infty } \frac{(-1)^{n+1} \ln^{2} n}{2^{n}};

(5)

∑n=1∞(−1)nn5arcsin⁡π4n\sum_{n = 1}^{\infty } \frac{(-1)^{n}}{\sqrt[5]{n}} \arcsin \frac{\pi }{4 n};

(6)

∑n=1∞cos⁡3n⋅arctg⁡n+1n3+2\sum_{n = 1}^{\infty } \cos^{3} n \cdot \operatorname {arctg} \frac{n+1}{n^{3}+2};

(7)

∑n=1∞n2+3n3+4nln⁡(1+(−1)nn)\sum_{n = 1}^{\infty } \sqrt{\frac{n^{2}+3}{n^{3}+4 n}} \ln \left(1+\frac{(-1)^{n}}{n}\right);

(8)

∑n=1∞n3sin⁡n⋅e−n\sum_{n = 1}^{\infty } n^{3} \sin n \cdot e^{-\sqrt{n}}.

Problem 2.15.2

Prove that the series converge absolutely.

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(1)

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

(2)

∑n=1∞(−1)n(2n)!!(n+1)n\sum_{n = 1}^{\infty } \frac{(-1)^{n}(2 n) ! !}{(n+1)^{n}}

(3)

∑n=1∞(−1)nln⁡2(n+1)nn+1\sum_{n = 1}^{\infty } \frac{(-1)^{n} \ln^{2}(n+1)}{n \sqrt{n+1}}

(4)

∑n=1∞(−1)n(n+1)/2⋅2n+n23n+n3\sum_{n = 1}^{\infty }(-1)^{n(n+1) / 2} \cdot \frac{2^{n}+n^{2}}{3^{n}+n^{3}};

(5)

∑n=1∞(−1)nsin⁡3nnln⁡(n+1)⋅ln⁡2(n+2)\sum_{n = 1}^{\infty } \frac{(-1)^{n} \sin 3 n}{n \ln (n+1) \cdot \ln^{2}(n+2)};

(6)

∑n=1∞(sin⁡nn23−sin⁡(sin⁡nn23))\sum_{n = 1}^{\infty }\left(\frac{\sin n}{\sqrt[3]{n^{2}}}-\sin \left(\frac{\sin n}{\sqrt[3]{n^{2}}}\right)\right)

(7)

∑n=1∞(1nsin⁡(1/n)−cos⁡1n)cos⁡πn\sum_{n = 1}^{\infty }\left(\frac{1}{n \sin (1 / n)}-\cos \frac{1}{n}\right) \cos \pi n;

(8)

∑n=1∞(−1)n(arctg⁡1n−arcsin⁡1n)\sum_{n = 1}^{\infty }(-1)^{n}\left(\operatorname {arctg} \frac{1}{\sqrt{n}}-\arcsin \frac{1}{\sqrt{n}}\right).

Problem 2.15.3

Investigate the series for convergence.

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(1)

∑n=1∞(−1)nn+13\sum_{n = 1}^{\infty } \frac{(-1)^{n}}{\sqrt[3]{n+1}};

(2)

∑n=1∞(−1)n+1ln⁡nn\sum_{n = 1}^{\infty } \frac{(-1)^{n+1} \ln n}{\sqrt{n}};

(3)

∑n=1∞(−1)n−1ln⁡ln⁡(n+2)ln⁡(n+1)\sum_{n = 1}^{\infty } \frac{(-1)^{n-1} \ln \ln (n+2)}{\ln (n+1)};

(4)

∑n=1∞(−1)nn(n+2)n+14\sum_{n = 1}^{\infty } \frac{(-1)^{n} n}{(n+2) \sqrt[4]{n+1}};

(5)

∑n=1∞cos⁡(π4+πn)sin⁡1n\sum_{n = 1}^{\infty } \cos \left(\frac{\pi }{4}+\pi n\right) \sin \frac{1}{n};

(6)

∑n=1∞(−1)n(1−cos⁡πn)\sum_{n = 1}^{\infty }(-1)^{n}\left(1-\cos \frac{\pi }{\sqrt{n}}\right);

(7)

∑n=1∞(−1)n+11n2+1n\sum_{n = 1}^{\infty }(-1)^{n+1} \frac{1}{\sqrt[n]{n^{2}+1}};

(8)

∑n=1∞(−1)nn+2n2+4arctg⁡πn\sum_{n = 1}^{\infty }(-1)^{n} \frac{n+2}{\sqrt{n^{2}+4}} \operatorname {arctg} \frac{\pi }{\sqrt{n}}.

Problem 2.15.4

Investigate the series for convergence.

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(1)

∑n=1∞(−1)n(n−1)/21n\sum_{n = 1}^{\infty }(-1)^{n(n-1) / 2} \frac{1}{\sqrt{n}};

(2)

∑n=1∞sin⁡nn5;\sum_{n = 1}^{\infty } \frac{\sin n}{\sqrt[5]{n}} ;

(3)

∑n=1∞cos⁡(n+π/4)ln⁡2(n+1)\sum_{n = 1}^{\infty } \frac{\cos (n+\pi / 4)}{\ln^{2}(n+1)};

(4)

∑n=1∞(−1)ncos⁡22nn\sum_{n = 1}^{\infty }(-1)^{n} \frac{\cos^{2} 2 n}{\sqrt{n}};

(5)

∑n=1∞(−1)n⋅sin⁡2(n/2)n+15\sum_{n = 1}^{\infty }(-1)^{n} \cdot \frac{\sin^{2}(n / 2)}{\sqrt[5]{n+1}};

(6)

∑n=1∞∑k=1n1kln⁡(n+2)cos⁡3n\sum_{n = 1}^{\infty } \frac{\sum_{k = 1}^{n} \frac{1}{k}}{\ln (n+2)} \cos 3 n;

(7)

∑n=1∞∑k=1n1k!n5sin⁡2n\sum_{n = 1}^{\infty } \frac{\sum_{k = 1}^{n} \frac{1}{k !}}{\sqrt[5]{n}} \sin 2 n;

(8)

∑n=1∞(−1)n(n−1)/21n+1(1+2n)2\sum_{n = 1}^{\infty }(-1)^{n(n-1) / 2} \frac{1}{\sqrt{n+1}}\left(1+\frac{2}{n}\right)^{2}.

Problem 2.15.5

Investigate the series for convergence and absolute convergence.

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(1)

∑n=1∞(−1)n−1n+13n+2\sum_{n = 1}^{\infty }(-1)^{n-1} \frac{\sqrt[3]{n+1}}{\sqrt{n}+2}

(2)

∑n=1∞(−1)nn22n3n+1\sum_{n = 1}^{\infty }(-1)^{n} \frac{n^{2} 2^{n}}{3^{n}+1}

(3)

∑n=1∞(−1)nnln⁡(n+1)\sum_{n = 1}^{\infty } \frac{(-1)^{n}}{n \ln (n+1)}

(4)

∑n=1∞(n+1)sin⁡2n\sum_{n = 1}^{\infty }(n+1) \sin 2 n

(5)

∑n=1∞(−1)nn(n+1)n+2tg⁡1n\sum_{n = 1}^{\infty } \frac{(-1)^{n} n}{(n+1) \sqrt{n+2}} \operatorname {tg} \frac{1}{\sqrt{n}};

(6)

∑n=1∞(−1)nnln⁡(n+1)ln⁡ln⁡(n+2)\sum_{n = 1}^{\infty } \frac{(-1)^{n}}{n \ln (n+1) \ln \ln (n+2)}.

Problem 2.15.6

Investigate the series for convergence and absolute convergence.

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(1)

∑n=1∞(−1)n+1(n+4)n2+13(2+n2+3)\sum_{n = 1}^{\infty } \frac{(-1)^{n+1}(n+4)}{\sqrt[3]{n^{2}+1}\left(2+\sqrt{n^{2}+3}\right)};

(2)

∑n=1∞cos⁡(πn+π6)ln⁡(1+2n)\sum_{n = 1}^{\infty } \cos \left(\pi n+\frac{\pi }{6}\right) \ln \left(1+\frac{2}{n}\right);

(3)

∑n=1∞(−1)n2n3+(−1)n−1\sum_{n = 1}^{\infty } \frac{(-1)^{n}}{2 \sqrt[3]{n}+(-1)^{n-1}};

(4)

∑n=1∞ln⁡2ln⁡(n+2)ln⁡(n+1)sin⁡πn4\sum_{n = 1}^{\infty } \frac{\ln^{2} \ln (n+2)}{\ln (n+1)} \sin \frac{\pi n}{4};

(5)

∑n=1∞(−1)n2+(−1)nnln⁡(n+1)\sum_{n = 1}^{\infty }(-1)^{n} \frac{2+(-1)^{n}}{n \ln (n+1)};

(6)

∑n=1∞cos⁡n⋅cos⁡(1/n)n4\sum_{n = 1}^{\infty } \frac{\cos n \cdot \cos (1 / n)}{\sqrt[4]{n}}

Problem 2.15.7

Investigate the series for convergence and absolute convergence.

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(1)

∑n=1∞ln⁡6nncos⁡πn6\sum_{n = 1}^{\infty } \frac{\ln^{6} n}{\sqrt{n}} \cos \frac{\pi n}{6};

(2)

∑n=1∞(−1)[n]n\sum_{n = 1}^{\infty } \frac{(-1)^{[\sqrt{n}]}}{n};

(3)

∑n=1∞(−1)nn2+2n+3−n2−2n+3n\sum_{n = 1}^{\infty }(-1)^{n} \frac{\sqrt{n^{2}+2 n+3}-\sqrt{n^{2}-2 n+3}}{n};

(4)

∑n=1∞cos⁡4nn\sum_{n = 1}^{\infty } \frac{\cos^{4} n}{n}

(5)

∑n=1∞sin⁡4n−3/8n+1\sum_{n = 1}^{\infty } \frac{\sin^{4} n-3 / 8}{\sqrt{n+1}};

(6)

∑n=1∞(−1)nn3/2+ln⁡3nn3/2ln⁡(n+1)\sum_{n = 1}^{\infty } \frac{(-1)^{n} n^{3 / 2}+\ln^{3} n}{n^{3 / 2} \ln (n+1)}.

Problem 2.15.8

Investigate the series for convergence.

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(1)

∑n=1∞sin⁡3nn\sum_{n = 1}^{\infty } \frac{\sin^{3} n}{\sqrt{n}};

(2)

∑n=1∞cos⁡52nln⁡(n+1)\sum_{n = 1}^{\infty } \frac{\cos^{5} 2 n}{\ln (n+1)}

(3)

∑n=1∞sin⁡nn+sin⁡n\sum_{n = 1}^{\infty } \frac{\sin n}{\sqrt{n}+\sin n};

(4)

∑n=1∞sin⁡(sin⁡nn3)\sum_{n = 1}^{\infty } \sin \left(\frac{\sin n}{\sqrt[3]{n}}\right);

(5)

∑n=1∞(e(cos⁡n)/n−cos⁡1n)\sum_{n = 1}^{\infty }\left(e^{(\cos n) / \sqrt{n}}-\cos \frac{1}{n}\right);

(6)

∑n=1∞(−1)nn+(−1)n/(2n)\sum_{n = 1}^{\infty } \frac{(-1)^{n}}{\sqrt{n}+(-1)^{n} /(2 \sqrt{n})};

(7)

∑n=1∞cos⁡nn2/3+(−1)n\sum_{n = 1}^{\infty } \frac{\cos n}{n^{2 / 3}+(-1)^{n}};

(8)

∑n=1∞sin⁡nncos⁡1n\sum_{n = 1}^{\infty } \frac{\sin n}{\sqrt{n}} \cos \frac{1}{n}

(9)

∑n=1∞(−1)nnarctg⁡n+1n3\sum_{n = 1}^{\infty }(-1)^{n} n \operatorname {arctg} \frac{n+1}{n^{3}};

(10)

∑n=1∞(nn−1)cos⁡n\sum_{n = 1}^{\infty }(\sqrt[n]{n}-1) \cos n;

(11)

∑n=1∞(−1)nsin⁡(n2−1−n)\sum_{n = 1}^{\infty }(-1)^{n} \sin \left(\sqrt{n^{2}-1}-n\right);

(12)

∑n=1∞(−1)nsh⁡n2+12n3−n2\sum_{n = 1}^{\infty }(-1)^{n} \operatorname {sh} \frac{n^{2}+1}{2 n^{3}-n^{2}}.

Problem 2.15.9

Investigate the series for convergence.

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(1)

∑n=1∞1n4cos⁡π(n2+1)n\sum_{n = 1}^{\infty } \frac{1}{\sqrt[4]{n}} \cos \frac{\pi \left(n^{2}+1\right)}{n};

(2)

∑n=1∞sin⁡(πn2+1)\sum_{n = 1}^{\infty } \sin \left(\pi \sqrt{n^{2}+1}\right);

(3)

∑n=1∞sin⁡(n+1/n)ln⁡ln⁡(n+1)\sum_{n = 1}^{\infty } \frac{\sin (n+1 / n)}{\ln \ln (n+1)};

(4)

∑n=1∞(−1)[ln⁡n]n\sum_{n = 1}^{\infty } \frac{(-1)^{[\ln n]}}{n};

(5)

∑n=1∞sin⁡n2\sum_{n = 1}^{\infty } \sin n^{2};

(6)

∑n=1∞sin⁡n⋅sin⁡n2n\sum_{n = 1}^{\infty } \frac{\sin n \cdot \sin n^{2}}{n}

(7)

∑n=1∞cos⁡nln⁡2(n+1)\sum_{n = 1}^{\infty } \frac{\cos n}{\ln^{2}(n+1)};

(8)

∑n=1∞cos⁡nn+ln⁡n\sum_{n = 1}^{\infty } \frac{\cos n}{n+\ln n}

(9)

∑n=1∞sin⁡nln⁡n+1tg⁡1n\sum_{n = 1}^{\infty } \frac{\sin n}{\ln n+1} \operatorname {tg} \frac{1}{\sqrt{n}};

(10)

∑n=1∞cos⁡narctg⁡ln⁡nn+1\sum_{n = 1}^{\infty } \cos n \operatorname {arctg} \frac{\ln n}{\sqrt{n+1}};

(11)

∑n=1∞(etg⁡1n−1)sin⁡nln⁡(n+1)\sum_{n = 1}^{\infty }\left(e^{\operatorname {tg} \frac{1}{n}-1}\right) \frac{\sin n}{\ln (n+1)};

(12)

∑n=1∞(−1)nln⁡cos⁡nn+2\sum_{n = 1}^{\infty }(-1)^{n} \ln \cos \frac{\sqrt{n}}{n+2}.

Problem 2.15.10

Let f(x)=Pˉm(x)/Qp(x)f(x) = \bar{P}_{m}(x) / Q_{p}(x), where

Pm(x)=xm+a1xm−1+…+am−1x+amQp(x)=xp+b1xp−1+…+bp−1x+bp \begin{gathered} P_{m}(x) = x^{m}+a_{1} x^{m-1}+\ldots +a_{m-1} x+a_{m} \\ Q_{p}(x) = x^{p}+b_{1} x^{p-1}+\ldots +b_{p-1} x+b_{p}\end{gathered}

are polynomials, with Qp(x)≠0Q_{p}(x) \neq 0 for x⩾1x \geqslant 1. Investigate the series ∑n=1∞(−1)nf(n)\sum_{n = 1}^{\infty }(-1)^{n} f(n) for convergence and absolute convergence.

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Problem 2.15.11

Investigate the series ∑n=3∞annln⁡(n+1)\sum_{n = 3}^{\infty } \frac{a_{n}}{n \ln (n+1)} for convergence, where

an={1, if n=5k+1 and n=5k+2,−1, if n=5k,n=5k−1,n=5k−2,k∈N. a_{n} = \begin{cases} 1, \text{ if } n = 5 k+1 \text{ and } n = 5 k+2, \\ -1, \text{ if } n = 5 k, n = 5 k-1, n = 5 k-2, \end{cases} \quad k \in N.
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Problem 2.15.12

Find all values of α\alpha for which the series:

  1. converges absolutely;

  2. converges conditionally.

?
(1)

∑n=1∞(−1)n−1nα\sum_{n = 1}^{\infty } \frac{(-1)^{n-1}}{n^{\alpha }};

(2)

∑n=1∞(−1)nn−α\sum_{n = 1}^{\infty } \frac{(-1)^{n}}{n-\alpha }

(3)

∑n=1∞(−1)nnα+n−1\sum_{n = 1}^{\infty } \frac{(-1)^{n}}{n^{\alpha +n^{-1}}};

(4)

∑n=1∞ln⁡(1+(−1)n−1(n+1)α)\sum_{n = 1}^{\infty } \ln \left(1+\frac{(-1)^{n-1}}{(n+1)^{\alpha }}\right);

(5)

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

(6)

∑n=1∞(−1)n−1(n+1+(−1)n)α\sum_{n = 1}^{\infty } \frac{(-1)^{n-1}}{\left(\sqrt{n+1}+(-1)^{n}\right)^{\alpha }}.

Problem 2.15.13

Find all values of α\alpha for which the series:

  1. converges absolutely;

  2. converges conditionally.

?
(1)

∑n=1∞(−1)[n]nα\sum_{n = 1}^{\infty } \frac{(-1)^{[n]}}{n^{\alpha }};

(2)

∑n=1∞cos⁡nnα\sum_{n = 1}^{\infty } \frac{\cos n}{n^{\alpha }};

(3)

∑n=1∞sin⁡2nln⁡2nnα\sum_{n = 1}^{\infty } \frac{\sin 2 n \ln^{2} n}{n^{\alpha }};

(4)

∑n=1∞(−1)n−1(1⋅3⋅5…(2n−1)2⋅4⋅6…(2n))α\sum_{n = 1}^{\infty }(-1)^{n-1}\left(\frac{1 \cdot 3 \cdot 5 \ldots (2 n-1)}{2 \cdot 4 \cdot 6 \ldots (2 n)}\right)^{\alpha }

(5)

∑n=1∞α(α−1)(α−2)…(α−(n−1))n!\sum_{n = 1}^{\infty } \frac{\alpha (\alpha -1)(\alpha -2) \ldots (\alpha -(n-1))}{n !}

(6)

∑n=1∞sin⁡nxnln⁡α(n+1),0<x<π\sum_{n = 1}^{\infty } \frac{\sin n x}{n \ln^{\alpha }(n+1)}, \quad 0 < x < \pi.

Problem 2.15.14

Find all values of α\alpha for which the series:

  1. converges absolutely;

  2. converges conditionally.

?
(1)

∑n=1∞(−1)nsin⁡2nαn\sum_{n = 1}^{\infty } \frac{(-1)^{n} \sin^{2 n} \alpha }{n};

(2)

∑n=1∞(−1)n−12ncos⁡2nαn\sum_{n = 1}^{\infty }(-1)^{n-1} \frac{2^{n} \cos^{2 n} \alpha }{\sqrt{n}};

(3)

1+13α−12α+15α+17α−14α+…1+\frac{1}{3^{\alpha }}-\frac{1}{2^{\alpha }}+\frac{1}{5^{\alpha }}+\frac{1}{7^{\alpha }}-\frac{1}{4^{\alpha }}+\ldots;

(4)

1+13α−11α+15α+17α−13α+19α+111α−15α+…1+\frac{1}{3^{\alpha }}-\frac{1}{1^{\alpha }}+\frac{1}{5^{\alpha }}+\frac{1}{7^{\alpha }}-\frac{1}{3^{\alpha }}+\frac{1}{9^{\alpha }}+\frac{1}{11^{\alpha }}-\frac{1}{5^{\alpha }}+\ldots

Problem 2.15.15

Find all values of α\alpha and β\beta for which the series:

  1. converges absolutely;

  2. converges conditionally:

?
(1)

∑n=1∞(−1)n−1(1+α)(2+α)…(n+α)nβn!\sum_{n = 1}^{\infty }(-1)^{n-1} \frac{(1+\alpha )(2+\alpha ) \ldots (n+\alpha )}{n^{\beta } n !};

(2)

11α−12β+13α−14β+15α−16β+…\frac{1}{1^{\alpha }}-\frac{1}{2^{\beta }}+\frac{1}{3^{\alpha }}-\frac{1}{4^{\beta }}+\frac{1}{5^{\alpha }}-\frac{1}{6^{\beta }}+\ldots;

(3)

1−22β+13α+14α−25β+16α+17α−28β+19α+…1-\frac{2}{2^{\beta }}+\frac{1}{3^{\alpha }}+\frac{1}{4^{\alpha }}-\frac{2}{5^{\beta }}+\frac{1}{6^{\alpha }}+\frac{1}{7^{\alpha }}-\frac{2}{8^{\beta }}+\frac{1}{9^{\alpha }}+\ldots.

Problem 2.15.16

Prove that the series ∑n=1∞sin⁡(xn)nα\sum_{n = 1}^{\infty } \frac{\sin (x \sqrt{n})}{n^{\alpha }} and ∑n=1∞cos⁡(xn)nα\sum_{n = 1}^{\infty } \frac{\cos (x \sqrt{n})}{n^{\alpha }}, where x≠0x \neq 0, converge for α>1/2\alpha > 1 / 2 and diverge for α<1/2\alpha < 1 / 2.

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Problem 2.15.17

Show that the series

1+13−12+15+17−14+19+111−16+… 1+\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{5}}+\frac{1}{\sqrt{7}}-\frac{1}{\sqrt{4}}+\frac{1}{\sqrt{9}}+\frac{1}{\sqrt{11}}-\frac{1}{\sqrt{6}}+\ldots

obtained from the convergent series ∑n=1∞(−1)n−1n\sum_{n = 1}^{\infty } \frac{(-1)^{n-1}}{\sqrt{n}} by rearranging its terms, diverges.

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Problem 2.15.18

All terms of the harmonic series ∑n=1∞1n\sum_{n = 1}^{\infty } \frac{1}{n} whose indices contain the digit 9 are discarded. Prove that the resulting series is convergent and its sum is less than 20.

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Problem 2.15.19

Using one of the equalities

1−12+13−14+…+12n−1−12n=1n+1+1n+2+…+12n1+12+13+…+1n=C+ln⁡n+εn \begin{gathered} 1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\ldots +\frac{1}{2 n-1}-\frac{1}{2 n} = \frac{1}{n+1}+\frac{1}{n+2}+\ldots +\frac{1}{2 n} \\ 1+\frac{1}{2}+\frac{1}{3}+\ldots +\frac{1}{n} = C+\ln n+\varepsilon _{n}\end{gathered}

where εn→0\varepsilon_{n} \rightarrow 0 as n→∞n \rightarrow \infty (see problem 16)), prove that

∑n=1∞(−1)n−1n=ln⁡2 \sum _{n = 1}^{\infty } \frac{(-1)^{n-1}}{n} = \ln 2
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Problem 2.15.20

Using the fact that the sum of the series ∑n=1∞(−1)n−1n\sum_{n = 1}^{\infty } \frac{(-1)^{n-1}}{n} equals ln⁡2\ln 2, find the sums of the following series, obtained from the given one by rearranging its terms:

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(1)

1+13−12+15+17−14+19+111−16+…1+\frac{1}{3}-\frac{1}{2}+\frac{1}{5}+\frac{1}{7}-\frac{1}{4}+\frac{1}{9}+\frac{1}{11}-\frac{1}{6}+\ldots;

(2)

1−12−14+13−16−18+15−…1-\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-\frac{1}{6}-\frac{1}{8}+\frac{1}{5}-\ldots;

(3)

1−12−14−16−18+13−110−112−114−116+15−…1-\frac{1}{2}-\frac{1}{4}-\frac{1}{6}-\frac{1}{8}+\frac{1}{3}-\frac{1}{10}-\frac{1}{12}-\frac{1}{14}-\frac{1}{16}+\frac{1}{5}-\ldots

Problem 2.15.21

Prove that if the terms of the series ∑n=1∞(−1)nn\sum_{n = 1}^{\infty } \frac{(-1)^{n}}{n} are rearranged so that each group of pp consecutive positive terms is followed by a group of mm consecutive negative terms, then the sum of the resulting series equals

ln⁡2+12ln⁡pm \ln 2+\frac{1}{2} \ln \frac{p}{m}
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Problem 2.15.22

Let the series ∑n=1∞un\sum_{n = 1}^{\infty } u_{n} be obtained from the series ∑n=1∞(−1)nn\sum_{n = 1}^{\infty } \frac{(-1)^{n}}{n} by rearranging its terms so that the terms of like sign are arranged in the new series in decreasing order of their absolute values, and the ratio of the number of positive summands in ∑k=1nuk\sum_{k = 1}^{n} u_{k} to the number of negative summands in this sum has, as n→∞n \rightarrow \infty, a limit equal to λ\lambda. Prove that

∑n=1∞un=ln⁡4λ \sum _{n = 1}^{\infty } u_{n} = \ln \sqrt{4 \lambda }
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Problem 2.15.23

Prove that the harmonic series ∑n=1∞1n\sum_{n = 1}^{\infty } \frac{1}{n} remains divergent if, without rearranging its terms, the signs of the terms are changed so that each group of pp positive terms is followed by a group of mm negative terms, where p≠mp \neq m. Show that for p=mp = m the resulting series is convergent.

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Problem 2.15.24

Let α>0\alpha > 0 and let SS be the sum of the series ∑n=1∞(−1)n−1nα\sum_{n = 1}^{\infty } \frac{(-1)^{n-1}}{n^{\alpha }}. Prove that 1/2<S<11 / 2 < S < 1.

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Problem 2.15.25

Let an>0(n∈N)a_{n} > 0(n \in N) and lim⁡n→∞an=0\lim_{n \rightarrow \infty } a_{n} = 0. Does it follow that the alternating series ∑n=1∞(−1)n−1an\sum_{n = 1}^{\infty }(-1)^{n-1} a_{n} converges?

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Problem 2.15.26

Let the series ∑n=1∞an\sum_{n = 1}^{\infty } a_{n} converge and lim⁡n→∞bnan=1\lim_{n \rightarrow \infty } \frac{b_{n}}{a_{n}} = 1. Does it follow that the series ∑n=1∞bn\sum_{n = 1}^{\infty } b_{n} also converges?

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Problem 2.15.27

Let the series ∑n=1∞an\sum_{n = 1}^{\infty } a_{n} and ∑n=1∞bn\sum_{n = 1}^{\infty } b_{n} converge and, for all n⩾n0n \geqslant n_{0}, the inequalities hold

an⩽cn⩽bn a_{n} \leqslant c_{n} \leqslant b_{n}

Prove that the series ∑n=1∞cn\sum_{n = 1}^{\infty } c_{n} converges.

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Problem 2.15.28

Prove that the series ∑n=1∞(−1)n−1an\sum_{n = 1}^{\infty }(-1)^{n-1} a_{n}, where an>0(n∈N)a_{n} > 0(n \in N), converges if there exists a number α>0\alpha > 0 such that

anan+1=1+αn+o(1n),n→∞ \frac{a_{n}}{a_{n+1}} = 1+\frac{\alpha }{n}+o\left(\frac{1}{n}\right), \quad n \rightarrow \infty
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Problem 2.15.29

Show that the sum of a non-absolutely convergent series does not change if the terms of this series are rearranged so that none of them is displaced from its former position by more than mm places, where mm is a given number.

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Problem 2.15.30

Show that the terms of a non-absolutely convergent series can be grouped, without rearrangement, so that the resulting series is absolutely convergent.

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Problem 2.15.31

Prove that the series ∑n=1∞anbn\sum_{n = 1}^{\infty } a_{n} b_{n} converges if the following conditions hold:

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(1)

the series ∑n=1∞bn\sum_{n = 1}^{\infty } b_{n} converges;

(2)

the series ∑n=1∞(an−an+1)\sum_{n = 1}^{\infty }\left(a_{n}-a_{n+1}\right) converges absolutely.

Problem 2.15.32

Prove that the series ∑n=1∞anbn\sum_{n = 1}^{\infty } a_{n} b_{n} converges if the following conditions hold:

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(1)

the partial sums of the series ∑n=1∞bn\sum_{n = 1}^{\infty } b_{n} are bounded, i.e., there exists a number M>0M > 0 such that for all n∈Nn \in N the inequality ∣∑k=1nbk∣⩽M\left|\sum_{k = 1}^{n} b_{k}\right| \leqslant M holds;

(2)

the series ∑n=1∞(an−an+1)\sum_{n = 1}^{\infty }\left(a_{n}-a_{n+1}\right) converges absolutely;

(3)

lim⁡n→∞an=0\lim_{n \rightarrow \infty } a_{n} = 0.

Problem 2.15.33

Let a numerical sequence {an}\left\{ a_{n}\right\} and a strictly increasing sequence of natural numbers {pk}\left\{ p_{k}\right\} be given such that

pk−pk−1<c p_{k}-p_{k-1} < c

cc is a given number, k∈Nk \in N. Denote

A1=∑j=1p1aj,A2=∑j=p1+1p2aj,…,Ak=∑j=pk−1+1pkaj(k⩾2). A_{1} = \sum _{j = 1}^{p_{1}} a_{j}, \quad A_{2} = \sum _{j = p_{1}+1}^{p_{2}} a_{j}, \quad \ldots , \quad A_{k} = \sum _{j = p_{k-1}+1}^{p_{k}} a_{j} \quad (k \geqslant 2).

Prove that if lim⁡n→∞an=0\lim_{n \rightarrow \infty } a_{n} = 0 and the series ∑k=1∞Ak\sum_{k = 1}^{\infty } A_{k} converges, then the series ∑n=1∞an\sum_{n = 1}^{\infty } a_{n} also converges.

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Problem 2.15.34

Let the series ∑n=1∞an\sum_{n = 1}^{\infty } a_{n} converge conditionally. Denote

αn=∣an∣+an2,βn=∣an∣−an2;An=∑k=1nαk,Bn=∑k=1nβk\alpha_{n} = \frac{\left|a_{n}\right|+a_{n}}{2}, \quad \beta_{n} = \frac{\left|a_{n}\right|-a_{n}}{2} ; \quad A_{n} = \sum_{k = 1}^{n} \alpha_{k}, \quad B_{n} = \sum_{k = 1}^{n} \beta_{k}.

Prove that:

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(1)

lim⁡n→∞αn=lim⁡n→∞βn=0\lim_{n \rightarrow \infty } \alpha_{n} = \lim_{n \rightarrow \infty } \beta_{n} = 0;

(2)

the series ∑n=1∞αn\sum_{n = 1}^{\infty } \alpha_{n} and ∑n=1∞βn\sum_{n = 1}^{\infty } \beta_{n} diverge;

(3)

lim⁡n→∞AnBn=1\lim_{n \rightarrow \infty } \frac{A_{n}}{B_{n}} = 1.

Problem 2.15.35

Let the series ∑n=1∞an\sum_{n = 1}^{\infty } a_{n} converge conditionally. Prove that the terms of this series can be rearranged so that for the resulting series ∑n=1∞a~n\sum_{n = 1}^{\infty } \tilde{a}_{n} we have

lim⁡n→∞∑k=1na~k=+∞ \lim _{n \rightarrow \infty } \sum _{k = 1}^{n} \tilde{a}_{k} = +\infty
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Problem 2.15.36

Let Sn=∑k=1nsin⁡kkS_{n} = \sum_{k = 1}^{n} \frac{\sin k}{k}, and let AnA_{n} and BnB_{n} be, respectively, the sums of all the positive and all the negative terms contained in SnS_{n}. Prove that

lim⁡n→∞BnAn=−1 \lim _{n \rightarrow \infty } \frac{B_{n}}{A_{n}} = -1
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