9.6E: Mazoezi ya Sehemu ya 9.6

Katika mazoezi ya 1 - 11, tumia mtihani wa uwiano ili uone kama kila mfululizo\(\displaystyle \sum^∞_{n=1}a_n\) hujiunga au hupungua. Hali kama mtihani uwiano ni inconclusive.

1)\(\displaystyle \sum_{n=1}^∞\frac{1}{n!}\)

Jibu

\(\displaystyle \lim_{n\to \infty}\frac{a_{n+1}}{a_n}=0.\)Inajiunga na Mtihani wa Uwiano.

2)\(\displaystyle \sum_{n=1}^∞\frac{10^n}{n!}\)

3)\(\displaystyle \sum_{n=1}^∞\frac{n^2}{2^n}\)

Jibu

\(\displaystyle \lim_{n\to \infty} \frac{a_{n+1}}{a_n}=\lim_{n\to \infty}\frac{1}{2}\left(\frac{n+1}{n}\right)^2=\frac{1}{2}<1.\)Inajiunga na Mtihani wa Uwiano.

4)\(\displaystyle \sum_{n=1}^∞\frac{n^{10}}{2^n}\)

5)\(\displaystyle \sum_{n=1}^∞\frac{(n!)^3}{(3n)!}\)

Jibu

\(\displaystyle \lim_{n\to \infty} \frac{a_{n+1}}{a_n}=\frac{1}{27}<1.\)Inajiunga na Mtihani wa Uwiano.

6)\(\displaystyle \sum_{n=1}^∞\frac{2^{3n}(n!)^3}{(3n)!}\)

7)\(\displaystyle \sum_{n=1}^∞\frac{(2n)!}{n^{2n}}\)

Jibu

\(\displaystyle \lim_{n\to \infty} \frac{a_{n+1}}{a_n}=\frac{4}{e^2}<1.\)Inajiunga na Mtihani wa Uwiano.

8)\(\displaystyle \sum_{n=1}^∞\frac{(2n)!}{(2n)^n}\)

9)\(\displaystyle \sum_{n=1}^∞\frac{n!}{(n/e)^n}\)

Jibu

\(\displaystyle \lim_{n\to \infty} \frac{a_{n+1}}{a_n}=1.\)Uwiano mtihani ni inconclusive.

10)\(\displaystyle \sum_{n=1}^∞\frac{(2n)!}{(n/e)^{2n}}\)

11)\(\displaystyle \sum_{n=1}^∞\frac{(2^nn!)^2}{(2n)^{2n}}\)

Jibu

\(\displaystyle \lim_{n\to \infty} \frac{a_n}{a_{n+1}}=\frac{1}{e^2}<1.\)Inajiunga na Mtihani wa Uwiano.

Katika mazoezi 12 - 21, tumia mtihani wa mizizi ili ueleze ikiwa\(\displaystyle \sum^∞_{n=1}a_n\) hujiunga, wapi\(a_n\) kama ifuatavyo.

12)\(\displaystyle a_k=\left(\frac{k−1}{2k+3}\right)^k\)

13)\(\displaystyle a_k=\left(\frac{2k^2−1}{k^2+3}\right)^k\)

Jibu

\(\displaystyle \lim_{k\to \infty} (a_k)^{1/k}=2>1.\)Inapungua na Mtihani wa Mizizi.

14)\(\displaystyle a_n=\frac{(\ln n)^{2n}}{n^n}\)

15)\(\displaystyle a_n=n/2^n\)

Jibu

\(\displaystyle \lim_{n\to \infty} (a_n)^{1/n}=1/2<1.\)Inajiunga na Mtihani wa Mizizi.

16)\(\displaystyle a_n=n/e^n\)

17)\(\displaystyle a_k=\frac{k^e}{e^k}\)

Jibu

\(\displaystyle \lim_{k\to \infty} (a_k)^{1/k}=1/e<1.\)Inajiunga na Mtihani wa Mizizi.

18)\(\displaystyle a_k=\frac{π^k}{k^π}\)

19)\(\displaystyle a_n=\left(\frac{1}{e}+\frac{1}{n}\right)^n\)

Jibu

\(\displaystyle \lim_{n\to \infty} a^{1/n}_n=\lim_{n\to \infty} \frac{1}{e}+\frac{1}{n}=\frac{1}{e}<1.\)Inajiunga na Mtihani wa Mizizi.

20)\(\displaystyle a_k=\frac{1}{(1+\ln k)^k}\)

21)\(\displaystyle a_n=\frac{(\ln(1+\ln n))^n}{(\ln n)^n}\)

Jibu

\(\displaystyle \lim_{n\to \infty} a^{1/n}_n= \lim_{n\to \infty} \frac{(\ln(1+\ln n))}{(\ln n)}=0\)kwa utawala wa L'Hôpital. Inajiunga na Mtihani wa Mizizi.

Katika mazoezi ya 22 - 28, tumia mtihani wa uwiano au mtihani wa mizizi kama inafaa ili kuamua kama mfululizo\(\displaystyle \sum_{k=1}^∞a_k\) na masharti yaliyotolewa\(a_k\) hujiunga, au useme ikiwa mtihani hauwezi kufungwa.

22)\(\displaystyle a_k=\frac{k!}{1⋅3⋅5⋯(2k−1)}\)

23)\(\displaystyle a_k=\frac{2⋅4⋅6⋯2k}{(2k)!}\)

Jibu

\(\displaystyle \lim_{k\to \infty} \frac{a_{k+1}}{a_k}= \lim_{k\to \infty} \frac{1}{2k+1}=0.\)Inajiunga na Mtihani wa Uwiano.

24)\(\displaystyle a_k=\frac{1⋅4⋅7⋯(3k−2)}{3^kk!}\)

25)\(\displaystyle a_n=\left(1−\frac{1}{n}\right)^{n^2}\)

Jibu

\(\displaystyle \lim_{n\to \infty} (a_n)^{1/n}=1/e.\)Inajiunga na Mtihani wa Mizizi.

26)\(\displaystyle a_k=\left(\frac{1}{k+1}+\frac{1}{k+2}+⋯+\frac{1}{2k}\right)^k\quad \Big(\) Dokezo: Linganisha\(a^{1/k}_k\) na\(\displaystyle ∫^{2k}_k\frac{dt}{t}.\Big)\)

27)\(\displaystyle a_k=\left(\frac{1}{k+1}+\frac{1}{k+2}+⋯+\frac{1}{3k}\right)^k\)

Jibu

\(\displaystyle \lim_{k\to \infty} a^{1/k}_k=\ln(3)>1.\)Inapungua na Mtihani wa Mizizi.

28)\(\displaystyle a_n=\left(n^{1/n}−1\right)^n\)

Katika mazoezi ya 29 - 30, tumia mtihani wa uwiano ili ueleze ikiwa\(\displaystyle \sum_{n=1}^∞a_n\) hujiunga, au ueleze ikiwa mtihani wa uwiano haujulikani.

29)\(\displaystyle \sum_{n=1}^∞\frac{3^{n^2}}{2^{n^3}}\)

Jibu

\(\displaystyle \lim_{n\to \infty} \left|\frac{a_{n+1}}{a_n}\right|= \lim_{n\to \infty} \frac{3^{2n+1}}{2^{3n^2+3n+1}}=0.\)Inajiunga na Mtihani wa Uwiano.

30)\(\displaystyle \sum_{n=1}^∞\frac{2^{n^2}}{n^nn!}\)

Katika mazoezi 31, tumia vipimo vya kulinganisha mizizi na kikomo ili ueleze ikiwa\(\displaystyle \sum_{n=1}^∞a_n\) hujiunga.

31)\(\displaystyle \sum_{n=1}^∞\frac{1}{x^n_n}\) mara\(x_{n+1}=\frac{1}{2}x_n+\dfrac{1}{x_n}, x_1=1\) (ladha: Kupata kikomo cha\({x_n}\).)

Jibu

Inajiunga na Mizizi Mtihani na Limit Ulinganisho Mtihani tangu\(\displaystyle \lim_{n\to \infty} x_n=\sqrt{2}\).

Katika mazoezi 32 - 43, tumia mtihani sahihi ili uone kama mfululizo unajiunga.

32)\(\displaystyle \sum_{n=1}^∞\frac{n+1}{n^3+n^2+n+1}\)

33)\(\displaystyle \sum_{n=1}^∞\frac{(−1)^{n+1}(n+1)}{n^3+3n^2+3n+1}\)

Jibu

Inajiunga kabisa na kikomo kulinganisha \(p\)na -mfululizo,\(p=2.\)

34)\(\displaystyle \sum_{n=1}^∞\frac{(n+1)^2}{n^3+(1.1)^n}\)

35)\(\displaystyle \sum_{n=1}^∞\frac{(n−1)^n}{(n+1)^n}\)

Jibu

\(\displaystyle \lim_{n→∞}a_n=1/e^2≠0\). Mfululizo hutofautiana na Mtihani wa Tofauti.

36)\(\displaystyle a_n=\left(1+\frac{1}{n^2}\right)^n\)\(\Big(\) Kidokezo:\(\left(1+\dfrac{1}{n^2}\right)^{n^2}≈e.\Big)\)

37)\(\displaystyle a_k=1/2^{\sin^2k}\)

Jibu

Masharti wala huwa na sifuri:\(a_k≥1/2,\) tangu\(\sin^2x≤1.\)

38)\(\displaystyle a_k=2^{−\sin(1/k)}\)

39)\(\displaystyle a_n=1/(^{n+2}_n)\) wapi\( (^n_k)=\frac{n!}{k!(n−k)!}\)

Jibu

\(a_n=\dfrac{2}{(n+1)(n+2)},\)ambayo hujiunga kwa kulinganisha \(p\)na -mfululizo kwa\(p=2\).

40)\(\displaystyle a_k=1/(^{2k}_k)\)

41)\(\displaystyle a_k=2^k/(^{3k}_k)\)

Jibu

\(a_k=\dfrac{2^k1⋅2⋯k}{(2k+1)(2k+2)⋯3k}≤(2/3)^k\)hujiunga kwa kulinganisha na mfululizo wa kijiometri.

42)\(\displaystyle a_k=\left(\frac{k}{k+\ln k}\right)^k\quad\Big(\) Kidokezo:\(a_k=\left(1+\dfrac{\ln k}{k}\right)^{−(k/\ln k)\ln k}≈e^{−\ln k}.\Big)\)

43)\(\displaystyle a_k=\left(\frac{k}{k+\ln k}\right)^{2k}\quad\Big(\) Kidokezo:\(a_k=\left(1+\dfrac{\ln k}{k}\right)^{−(k/\ln k)\ln k^2}.\Big)\)

Jibu

\(a_k≈e^{−\ln k^2}=1/k^2.\)Mfululizo hujiunga na kikomo kulinganisha \(p\)na -mfululizo,\(p=2.\)

Mfululizo katika mazoezi 44 - 47 hujiunga na mtihani wa uwiano. Tumia summation kwa sehemu,\(\displaystyle \sum_{k=1}^na_k(b_{k+1}−b_k)=[a_{n+1}b_{n+1}−a_1b_1]−\sum_{k=1}^nb_{k+1}(a_{k+1}−a_k),\) ili kupata jumla ya mfululizo uliopewa.

44)\(\displaystyle \sum_{k=1}^∞\frac{k}{2^k}\) (kidokezo: Chukua\(a_k=k\) na\(b_k=2^{1−k}\).)

45) mara\(\displaystyle \sum_{k=1}^∞\frac{k}{c^k},\)\(c>1\) (dokezo: Chukua\(a_k=k\) na\(b_k=c^{1−k}/(c−1)\).)

Jibu

Ikiwa\(b_k=c^{1−k}/(c−1)\) na\(a_k=k\), basi\(b_{k+1}−b_k=−c^{−k}\) na\(\displaystyle \sum_{n=1}^∞\frac{k}{c^k}=a_1b_1+\frac{1}{c−1}\sum_{k=1}^∞c^{−k}=\frac{c}{(c−1)^2}.\)

46)\(\displaystyle \sum_{n=1}^∞\frac{n^2}{2^n}\)

47)\(\displaystyle \sum_{n=1}^∞\frac{(n+1)^2}{2^n}\)

Jibu

\(\displaystyle 6+4+1=11\)

\(k^{\text{th}}\)Neno la kila mfululizo wafuatayo lina sababu\(x^k\). Pata\(x\) upeo ambao mtihani wa uwiano unamaanisha kuwa mfululizo unajiunga.

48)\(\displaystyle \sum_{k=1}^∞\frac{x^k}{k^2}\)

49)\(\displaystyle \sum_{k=1}^∞\frac{x^{2k}}{k^2}\)

Jibu

\( |x|≤1\)

50)\(\displaystyle \sum_{k=1}^∞\frac{x^{2k}}{3^k}\)

51)\(\displaystyle \sum_{k=1}^∞\frac{x^k}{k!}\)

Jibu

\( |x|<∞\)

52) Je, kuna idadi kama\(p\) hiyo\(\displaystyle \sum_{n=1}^∞\frac{2^n}{n^p}\) inayogeuka?

53) Hebu\( 0<r<1.\) Kwa idadi\(p\) gani halisi\(\displaystyle \sum_{n=1}^∞n^pr^n\) inayojiunga?

Jibu

Nambari zote halisi\(p\) na Mtihani wa Uwiano.

54) Tuseme kwamba\(\displaystyle \lim_{n→∞}\left|\frac{a_{n+1}}{a_n}\right|=p.\) Kwa maadili gani ya\(p\) lazima\(\displaystyle \sum_{n=1}^∞2^na_n\) yanajiunga?

55) Tuseme kwamba\(\displaystyle \lim_{n→∞}\left|\frac{a_{n+1}}{a_n}\right|=p.\) Kwa maadili gani ya\(r>0\) ni\(\displaystyle \sum_{n=1}^∞r^na_n\) uhakika wa kuungana?

Jibu

\( r<1/p\)

56) Tuseme kwamba\(\left|\dfrac{a_{n+1}}{a_n}\right| ≤(n+1)^p\) kwa wote\(n=1,2,…\) ambapo\(p\) ni fasta halisi idadi. Kwa maadili gani ya\(p\) ni\(\displaystyle \sum_{n=1}^∞n!a_n\) uhakika wa kugeuza?

57) Kwa maadili gani ya\(r>0\), ikiwa ni yoyote,\(\displaystyle \sum_{n=1}^∞r^{\sqrt{n}}\) yanajiunga? \(\Big(\)Kidokezo:\(\displaystyle sum_{n=1}^∞a_n=\sum_{k=1}^∞\sum_{n=k^2}^{(k+1)^2−1}a_n.\Big)\)

Jibu

\(0<r<1.\)Kumbuka kuwa vipimo vya uwiano na mizizi havikubaliki. Kutumia ladha, kuna\(2k\) maneno\(r^\sqrt{n}\)\( k^2≤n<(k+1)^2\), na kwa\(r<1\) kila neno ni angalau\(r^k\). Hivyo,\(\displaystyle \sum_{n=1}^∞r^{\sqrt{n}}=\sum_{k=1}^∞\sum_{n=k^2}^{(k+1)^2−1}r^{\sqrt{n}} ≥\sum_{k=1}^∞2kr^k,\) ambayo hujiunga na mtihani wa uwiano kwa\(r<1\). Kwa\(r≥1\) mfululizo hutofautiana na mtihani wa tofauti.

58) Tuseme kwamba\( \left|\dfrac{a_{n+2}}{a_n}\right| ≤r<1\) kwa wote\(n\). Je, unaweza kuhitimisha kwamba\(\displaystyle \sum_{n=1}^∞a_n\) converges?

59) Hebu\(a_n=2^{−[n/2]}\)\( [x]\) wapi integer kubwa chini ya au sawa na\(x\). Kuamua kama\(\displaystyle \sum_{n=1}^∞a_n\) hujiunga na kuhalalisha jibu lako.

Jibu

Mmoja ana\(\displaystyle a_1=1, a_2=a_3=1/2,…a_{2n}=a_{2n+1}=1/2^n\). Mtihani wa uwiano hautumiki kwa sababu\(\displaystyle n\) ni\(\displaystyle a_{n+1}/a_n=1\) hata. Hata hivyo,\(\displaystyle a_{n+2}/a_n=1/2,\) hivyo mfululizo hujiunga kulingana na zoezi la awali. Bila shaka, mfululizo ni mfululizo wa kijiometri tu.

Mazoezi yafuatayo ya juu hutumia mtihani wa uwiano wa jumla ili kuamua muunganiko wa mfululizo fulani unaojitokeza katika maombi fulani wakati vipimo katika sura hii, ikiwa ni pamoja na uwiano na mtihani wa mizizi, hawana nguvu ya kutosha kuamua muunganiko wao. Jaribio linasema kwamba ikiwa\(\displaystyle \lim_{n→∞}\frac{a_{2n}}{a_n}<1/2\), basi\(\displaystyle \sum a_n\) hujiunga, wakati ikiwa\(\displaystyle \lim_{n→∞}\frac{a_{2n+1}}{a_n}>1/2\), basi\(\displaystyle \sum a_n\) hutofautiana.

60) Hebu\(\displaystyle a_n=\frac{1}{4}\frac{3}{6}\frac{5}{8}⋯\frac{2n−1}{2n+2}=\frac{1⋅3⋅5⋯(2n−1)}{2^n(n+1)!}\). Eleza kwa nini mtihani uwiano hauwezi kuamua muunganiko wa\(\displaystyle \sum_{n=1}^∞a_n\). Tumia ukweli\(\displaystyle 1−1/(4k)\) unaoongezeka\(\displaystyle k\) ili kukadiria\(\displaystyle \lim_{n→∞}\frac{a_{2n}}{a_n}\).

61) Hebu\(\displaystyle a_n=\frac{1}{1+x}\frac{2}{2+x}⋯\frac{n}{n+x}\frac{1}{n}=\frac{(n−1)!}{(1+x)(2+x)⋯(n+x).}\) Onyesha kwamba\(a_{2n}/a_n≤e^{−x/2}/2\). Kwa\(x>0\) nini mtihani wa uwiano wa jumla unamaanisha kuunganishwa kwa\(\displaystyle \sum_{n=1}^∞a_n\)? (Kidokezo: Andika\(2a_{2n}/a_n\) kama bidhaa ya\(n\) mambo ya kila ndogo kuliko\(1/(1+x/(2n)).)\)

Jibu

\(\displaystyle a_{2n}/a_n=\frac{1}{2}⋅\frac{n+1}{n+1+x}\frac{n+2}{n+2+x}⋯\frac{2n}{2n+x}.\)Inverse ya\(\displaystyle kth\) sababu ni\(\displaystyle (n+k+x)/(n+k)>1+x/(2n)\) hivyo bidhaa ni chini ya\(\displaystyle (1+x/(2n))^{−n}≈e^{−x/2}.\) Hivyo kwa\(\displaystyle x>0, \frac{a_{2n}}{a_n}≤\frac{1}{2}e^{−x/2}\). mfululizo hujiunga kwa\(\displaystyle x>0\).

62) Hebu\(a_n=\dfrac{n^{\ln n}}{(\ln n)^n}.\) Onyesha kwamba\( \dfrac{a_{2n}}{a_n}→0\) kama\(n→∞.\)