9.4E: Mazoezi ya Sehemu ya 9.4

Tumia Mtihani wa Kulinganisha ili kuamua kama kila mfululizo katika mazoezi ya 1 - 13 hujiunga au hupungua.

1)\(\displaystyle \sum^∞_{n=1}a_n\) wapi\(a_n=\dfrac{2}{n(n+1)}\)

2)\(\displaystyle \sum^∞_{n=1}a_n\) wapi\(a_n=\dfrac{1}{n(n+1/2)}\)

Jibu

Inajiunga kwa kulinganisha na\(\dfrac{1}{n^2}\).

3)\(\displaystyle \sum^∞_{n=1}\frac{1}{2(n+1)}\)

4)\(\displaystyle \sum^∞_{n=1}\frac{1}{2n−1}\)

Jibu

Inapungua kwa kulinganisha na mfululizo wa harmonic, tangu\(2n−1≥n.\)

5)\(\displaystyle \sum^∞_{n=2}\frac{1}{(n\ln n)^2}\)

6)\(\displaystyle \sum^∞_{n=1}\frac{n!}{(n+2)!}\)

Jibu

\(a_n=1/(n+1)(n+2)<1/n^2.\)Inajiunga kwa kulinganisha na\(p\) -mfululizo,\(p=2>1\).

7)\(\displaystyle \sum^∞_{n=1}\frac{1}{n!}\)

8)\(\displaystyle \sum^∞_{n=1}\frac{\sin(1/n)}{n}\)

Jibu

\(\sin(1/n)≤1/n,\)hivyo hujiunga kwa kulinganisha na\(p\) -mfululizo,\(p=2>1\).

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

10)\(\displaystyle \sum_{n=1}^∞\frac{\sin(1/n)}{\sqrt{n}}\)

Jibu

\(\sin(1/n)≤1,\)hivyo hujiunga kwa kulinganisha na\(p\) -mfululizo,\(p=3/2>1.\)

11)\(\displaystyle \sum^∞_{n=1}\frac{n^{1.2}−1}{n^{2.3}+1}\)

12)\(\displaystyle \sum^∞_{n=1}\frac{\sqrt{n+1}−\sqrt{n}}{n}\)

Jibu

Kwa kuwa\(\sqrt{n+1}−\sqrt{n}=1/(\sqrt{n+1}+\sqrt{n})≤2/\sqrt{n},\) mfululizo hujiunga kwa kulinganisha na\(p\) -mfululizo kwa\(p=1.5>1\).

13)\(\displaystyle \sum^∞_{n=1}\frac{\sqrt[4]{n}}{\sqrt[3]{n^4+n^2}}\)

Tumia Mtihani wa Ulinganisho wa Limit kuamua kama kila mfululizo katika mazoezi 14 - 28 hujiunga au hupungua.

14)\(\displaystyle \sum^∞_{n=1}\left(\frac{\ln n}{n}\right)^2\)

Jibu

Inajiunga na kikomo kulinganisha na\(p\) -mfululizo kwa\(p>1\).

15)\(\displaystyle \sum^∞_{n=1}\left(\frac{\ln n}{n^{0.6}}\right)^2\)

16)\(\displaystyle \sum^∞_{n=1}\frac{\ln\left(1+\frac{1}{n}\right)}{n}\)

Jibu

Inajiunga na kikomo kulinganisha na\(p\) -mfululizo,\(p=2>1.\)

17)\(\displaystyle \sum^∞_{n=1}\ln\left(1+\frac{1}{n^2}\right)\)

18)\(\displaystyle \sum^∞_{n=1}\frac{1}{4^n−3^n}\)

Jibu

Inajiunga na kikomo kulinganisha na\(4^{−n}\).

19)\(\displaystyle \sum^∞_{n=1}\frac{1}{n^2−n\sin n}\)

20)\(\displaystyle \sum^∞_{n=1}\frac{1}{e^{(1.1)n}−3^n}\)

Jibu

Inajiunga na kikomo kulinganisha na\(1/e^{1.1n}\).

21)\(\displaystyle \sum^∞_{n=1}\frac{1}{e^{(1.01)n}−3^n}\)

22)\(\displaystyle \sum^∞_{n=1}\frac{1}{n^{1+1/n}}\)

Jibu

Inapingana na kikomo kulinganisha na mfululizo harmonic.

23)\(\displaystyle \sum^∞_{n=1}\frac{1}{2^{1+1/n}}{n^{1+1/n}}\)

24)\(\displaystyle \sum^∞_{n=1}\left(\frac{1}{n}−\sin\left(\frac{1}{n}\right)\right)\)

Jibu

Inajiunga na kikomo kulinganisha na\(p\) -mfululizo,\(p=3>1\).

25)\(\displaystyle \sum^∞_{n=1}\left(1−\cos\left(\frac{1}{n}\right)\right)\)

26)\(\displaystyle \sum^∞_{n=1}\frac{1}{n}\left(\tan^{−1}n−\frac{π}{2}\right)\)

Jibu

Inajiunga na kikomo kulinganisha na\(p\) -mfululizo,\(p=3>1\).

27)\(\displaystyle \sum^∞_{n=1}\left(1−\frac{1}{n}\right)^{n.n}\) (Kidokezo:\(\left(1−\dfrac{1}{n}\right)^n→1/e.\))

28)\(\displaystyle \sum^∞_{n=1}\left(1−e^{−1/n}\right)\) (Kidokezo:\(1/e≈(1−1/n)^n,\) hivyo\(1−e^{−1/n}≈1/n.\))

Jibu

Inapingana na kikomo kulinganisha na\(1/n\).

29) Je,\(\displaystyle \sum^∞_{n=2}\frac{1}{(\ln n)^p}\) hujiunga ikiwa\(p\) ni kubwa ya kutosha? Kama ni hivyo, kwa ajili ya ambayo\(p?\)

30) Je,\(\displaystyle \sum^∞_{n=1}\left(\frac{\ln n}{n}\right)^p\) hujiunga ikiwa\(p\) ni kubwa ya kutosha? Kama ni hivyo, kwa ajili ya ambayo\(p?\)

Jibu

Inajiunga\(p>1\) kwa kulinganisha na\(p\) mfululizo kwa ndogo kidogo\(p\).

31) Kwa\(p\) nini mfululizo\(\displaystyle \sum^∞_{n=1}2^{pn}/3^n\) hujiunga?

32) Kwa\(p>0\) nini mfululizo\(\displaystyle \sum^∞_{n=1}\frac{n^p}{2^n}\) hujiunga?

Jibu

Inajiunga kwa ajili ya wote\(p>0\).

33) Kwa\(r>0\) nini mfululizo\(\displaystyle \sum^∞_{n=1}\frac{r^{n^2}}{2^n}\) hujiunga?

34) Kwa\(r>0\) nini mfululizo\(\displaystyle \sum^∞_{n=1}\frac{2^n}{r^{n^2}}\) hujiunga?

Jibu

Inajiunga kwa ajili ya wote\(r>1\). Ikiwa\(r>1\) basi\(r^n>4\), sema, mara moja\(n>\ln(2)/\ln(r)\) na kisha mfululizo hujiunga na kikomo kulinganisha na mfululizo wa kijiometri na uwiano\(1/2\).

35) Kupata maadili yote ya\(p\) na\(q\) kama kwamba\(\displaystyle \sum^∞_{n=1}\frac{n^p}{(n!)^q}\) converges.

36) Je,\(\displaystyle \sum^∞_{n=1}\frac{\sin^2(nr/2)}{n}\) hujiunga au kutofautiana? Eleza.

Jibu

Nambari ni sawa na\(1\) wakati\(n\) ni isiyo ya kawaida na\(0\) wakati\(n\) ni hata, hivyo mfululizo unaweza kuandikwa upya\(\displaystyle \sum^∞_{n=1}\frac{1}{2n+1},\) ambayo diverges kwa kikomo kulinganisha na mfululizo harmonic.

37) Eleza kwa nini\(n\), kwa kila, angalau moja ya\({|\sin n|,|\sin(n+1)|,...,|\sin(n+6)|}\) ni kubwa kuliko\(1/2\). Kutumia uhusiano huu kwa mtihani muunganiko wa\(\displaystyle \sum^∞_{n=1}\frac{|\sin n|}{\sqrt{n}}\).

38) Tuseme kwamba\(a_n≥0\)\(b_n≥0\) na kwamba\(\displaystyle \sum_{n=1}^∞a^2_n\) na\(\displaystyle \sum_{n=1}^∞b^2_n\) kuungana. Thibitisha kwamba\(\displaystyle \sum_{n=1}^∞a_nb_n\) hujiunga na\(\displaystyle \sum_{n=1}^∞a_nb_n≤\frac{1}{2}\left(\sum_{n=1}^∞a^2_n+\sum_{n=1}^∞b^2_n\right)\).

Jibu

\((a−b)^2=a^2−2ab+b^2\)au\(a^2+b^2≥2ab\), hivyo muunganiko ifuatavyo kutoka kulinganisha\(2a_nb_n\) na\(a^2_n+b^2_n.\) Tangu kiasi cha sehemu upande wa kushoto ni imepakana na wale wa kulia, usawa ana kwa mfululizo usio.

39) Je,\(\displaystyle \sum_{n=1}^∞2^{−\ln\ln n}\) hujiunga? (kidokezo: Andika\(2^{\ln\ln n}\) kama nguvu ya\(\ln n\).)

40) Je,\(\displaystyle \sum_{n=1}^∞(\ln n)^{−\ln n}\) hujiunga? (Kidokezo: Tumia\(t=e^{\ln(t)}\) kulinganisha na\(p\) mfululizo wa -.)

Jibu

\((\ln n)^{−\ln n}=e^{−\ln(n)\ln\ln(n)}.\)Ikiwa\(n\) ni kubwa ya kutosha, basi\(\ln\ln n>2,\) hivyo\((\ln n)^{−\ln n}<1/n^2\), na mfululizo hujiunga kwa kulinganisha na\(p\) mfululizo wa -.

41) Je,\(\displaystyle \sum_{n=2}^∞(\ln n)^{−\ln\ln n}\) hujiunga? (Kidokezo: Linganisha\(a_n\) na\(1/n\).)

42) Onyesha kwamba ikiwa\(a_n≥0\) na\(\displaystyle \sum_{n=1}^∞a_n\) hujiunga, kisha\(\displaystyle \sum_{n=1}^∞a^2_n\) hujiunga. Ikiwa\(\displaystyle \sum_{n=1}^∞a^2_n\) hujiunga, je,\(\displaystyle \sum_{n=1}^∞a_n\) lazima hujiunga?

Jibu

\(a_n→0,\)hivyo\(a^2_n≤|a_n|\) kwa ajili ya kubwa\(n\). Convergence ifuatavyo kutoka kikomo kulinganisha. \(\displaystyle \sum_{n=1}^∞\frac{1}{n^2}\)converges, lakini\(\displaystyle \sum_{n=1}^∞\frac{1}{n}\) hana, hivyo ukweli kwamba\(\displaystyle \sum_{n=1}^∞a^2_n\) converges haimaanishi kwamba\(\displaystyle \sum_{n=1}^∞a_n\) converges.

43) Tuseme kwamba\(a_n>0\) kwa wote\(n\) na kwamba\(\displaystyle \sum_{n=1}^∞a_n\) hujiunga. Tuseme kwamba\(b_n\) ni mlolongo wa kiholela wa zero na wale. Je,\(\displaystyle \sum_{n=1}^∞a_nb_n\) lazima hujiunga?

44) Tuseme kwamba\(a_n>0\) kwa wote\(n\) na kwamba\(\displaystyle \sum_{n=1}^∞a_n\) diverges. Tuseme kwamba\(b_n\) ni mlolongo wa kiholela wa zero na wale walio na maneno mengi sana sawa na moja. Je,\(\displaystyle \sum_{n=1}^∞a_nb_n\) lazima kutofautiana?

Jibu

Hapana. \(\displaystyle \sum_{n=1}^∞1/n\)hutengana. Hebu\(b_k=0\) isipokuwa\(k=n^2\) kwa baadhi\(n\). Kisha\(\displaystyle \sum_kb_k/k=\sum1/k^2\) hujiunga.

45) Kukamilisha maelezo ya hoja ifuatayo: Ikiwa\(\displaystyle \sum_{n=1}^∞\frac{1}{n}\) hujiunga na jumla ya mwisho\(s\), basi\(\dfrac{1}{2}s=\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{6}+⋯\) na\(s−\dfrac{1}{2}s=1+\dfrac{1}{3}+\dfrac{1}{5}+⋯.\) Kwa nini hii inasababisha utata?

46) Onyesha kwamba ikiwa\(a_n≥0\) na\(\displaystyle \sum_{n=1}^∞a^2_n\) hujiunga, kisha\(\displaystyle \sum_{n=1}^∞\sin^2(a_n)\) hujiunga.

Jibu

\(|\sin t|≤|t|,\)hivyo matokeo hufuata kutokana na mtihani wa kulinganisha.

47) Tuseme kwamba\(a_n/b_n→0\) katika mtihani wa kulinganisha, wapi\(a_n≥0\) na\(b_n≥0\). Thibitisha kwamba ikiwa\(\displaystyle \sum b_n\) hujiunga, kisha\(\displaystyle \sum a_n\) hujiunga.

48) Hebu\(b_n\) kuwa mlolongo usio na kipimo wa zero na wale. ni kubwa iwezekanavyo thamani gani ya\(\displaystyle x=\sum_{n=1}^∞b_n/2^n\)?

Jibu

Kwa mtihani wa kulinganisha,\(\displaystyle x=\sum_{n=1}^∞b_n/2^n≤\sum_{n=1}^∞1/2^n=1.\)

49) Hebu\(d_n\) kuwa mlolongo usio wa tarakimu, maana\(d_n\) inachukua maadili katika\(\{0,1,…,9\}\). ni kubwa iwezekanavyo thamani ya\(\displaystyle x=\sum_{n=1}^∞d_n/10^n\) kwamba converges nini?

50) Eleza kwa nini, ikiwa\(x>1/2,\) basi\(x\) haiwezi kuandikwa\(\displaystyle x=\sum_{n=2}^∞\frac{b_n}{2^n}\quad (b_n=0\;\text{or}\;1,\;b_1=0).\)

Jibu

Kama\(b_1=0,\) basi, kwa kulinganisha,\(\displaystyle x≤\sum_{n=2}^∞1/2^n=1/2.\)

51) [T] Evelyn ina kiwango cha kusawazisha kamili, idadi isiyo na ukomo wa uzito wa\(1\) kilo, na moja ya kila\(1/2\) moja-kg,\(1/4\) -kg,\(1/8\) -kg, na kadhalika uzito. Anataka kupima meteorite ya asili isiyojulikana kwa usahihi wa kiholela. Kutokana wadogo ni kubwa ya kutosha, anaweza kufanya hivyo? Je, hii inahusiana na mfululizo usio na mwisho?

52) [T] Robert anataka kujua mwili wake molekuli kwa usahihi holela. Ana kubwa kusawazisha wadogo kwamba kazi kikamilifu, ukusanyaji ukomo wa uzito\(1\) -kg, na tisa kila moja ya\(0.1\) -kg,\(0.01\) -kg,\(0.001\) -kg, na kadhalika uzito. Kutokana wadogo ni kubwa ya kutosha, anaweza kufanya hivyo? Je, hii inahusiana na mfululizo usio na mwisho?

Jibu

Ndiyo. Weka kuongeza uzito wa\(1\) kilo hadi vidokezo vya usawa kwa upande na uzito. Kama ni mizani kikamilifu, na Robert amesimama upande wa pili, kuacha. Vinginevyo, kuondoa moja ya uzito\(1\) -kg, na kuongeza uzito\(0.1\) -kg moja kwa wakati. Kama ni mizani baada ya kuongeza baadhi ya hizi, kuacha. Vinginevyo ikiwa ni vidokezo kwa uzito, ondoa uzito wa mwisho wa\(0.1\) kilo. Anza kuongeza uzito wa\(0.01\) kilo. Ikiwa ni mizani, simama. Ikiwa ni vidokezo kwa upande na uzito, ondoa uzito wa mwisho wa\(0.01\) kilo ulioongezwa. Endelea kwa njia hii kwa uzito\(0.001\) -kg, na kadhalika. Baada ya idadi ya mwisho ya hatua, mtu ana mfululizo wa mwisho wa fomu\(\displaystyle A+\sum_{n=1}^Ns_n/10^n\) ambapo\(A\) ni idadi ya uzito kamili wa kilo na\(d_n\) ni idadi ya uzito wa\(1/10^n\) kilo zilizoongezwa. Ikiwa katika hali fulani mfululizo huu ni uzito halisi wa Robert, mchakato utaacha. Vinginevyo inawakilisha jumla ya\(N^{\text{th}}\) sehemu ya mfululizo usio na kipimo ambayo inatoa uzito halisi wa Robert, na kosa la jumla hii ni zaidi\(1/10^N\).

53) Mfululizo\(\displaystyle \sum_{n=1}^∞\frac{1}{2n}\) ni nusu ya mfululizo wa harmonic na hivyo hutofautiana. Inapatikana kutoka mfululizo wa harmonic kwa kufuta maneno yote ambayo\(n\) ni isiyo ya kawaida. Hebu\(m>1\) iwe fasta. Onyesha, kwa ujumla, kwamba kufuta maneno yote\(1/n\) ambapo\(n=mk\) kwa integer fulani\(k\) pia husababisha mfululizo tofauti.

54) Kwa mtazamo wa zoezi la awali, inaweza kuwa ya kushangaza kwamba sehemu ndogo ya mfululizo wa harmonic ambayo karibu moja katika kila maneno tano hufutwa inaweza kugeuka. wazi harmonic mfululizo ni mfululizo kupatikana kutoka\(\displaystyle \sum_{n=1}^∞\frac{1}{n}\) kwa kuondoa muda wowote\(1/n\) kama tarakimu fulani, kusema\(9\), inaonekana katika upanuzi decimal ya\(n\). Wanasema kuwa mfululizo huu wa harmonic ulioharibika hujiunga na kujibu maswali yafuatayo.

a Nambari ngapi zote\(n\) zina\(d\) tarakimu?

b Ni nambari\(d\) ngapi za tarakimu\(h(d)\) nzima. hazina\(9\) kama moja au zaidi ya tarakimu zao?

c Nambari ndogo ya\(d\) tarakimu ni nini\(m(d)\)?

d. kueleza kwa nini ilifutwa harmonic mfululizo imepakana na\(\displaystyle \sum_{d=1}^∞\frac{h(d)}{m(d)}\).

e Onyesha kwamba\(\displaystyle \sum_{d=1}^∞\frac{h(d)}{m(d)}\) hujiunga.

Jibu

a.\(10^d−10^{d−1}<10^d\)
b.\(h(d)<9^d\)
c.\(m(d)=10^{d−1}+1\)
d. kundi la maneno katika mfululizo wa harmonic uliofutwa pamoja na idadi ya tarakimu. \(h(d)\)mipaka idadi ya maneno, na kila neno ni saa zaidi\(\frac{1}{m(d)}.\)
Kisha\(\displaystyle \sum_{d=1}^∞h(d)/m(d)≤\sum_{d=1}^∞9^d/(10)^{d−1}≤90\). Mtu anaweza kweli kutumia kulinganisha kukadiria thamani ndogo kuliko\(80\). Thamani halisi ni ndogo kuliko\(23\).

55) Tuseme kwamba mlolongo wa idadi\(a_n>0\) ina mali ambayo\(a_1=1\) na\(a_{n+1}=\dfrac{1}{n+1}S_n\), ambapo\(S_n=a_1+⋯+a_n\). Je, unaweza kuamua kama\(\displaystyle \sum_{n=1}^∞a_n\) converges? (Kidokezo:\(S_n\) ni monotone.)

56) Tuseme kwamba mlolongo wa idadi\(a_n>0\) ina mali ambayo\(a_1=1\) na\(a_{n+1}=\dfrac{1}{(n+1)^2}S_n\), ambapo\(S_n=a_1+⋯+a_n\). Je, unaweza kuamua kama\(\displaystyle \sum_{n=1}^∞a_n\) converges? (Kidokezo:\(S_2=a_2+a_1=a_2+S_1=a_2+1=1+1/4=(1+1/4)S_1, S_3=\dfrac{1}{3^2}S_2+S_2=(1+1/9)S_2=(1+1/9)(1+1/4)S_1\), nk Angalia\(\ln(S_n)\), na utumie\(\ln(1+t)≤t, t>0.\))

Jibu

Kuendelea ladha anatoa\(S_N=(1+1/N^2)(1+1/(N−1)^2…(1+1/4)).\) Kisha\(\ln(S_N)=\ln(1+1/N^2)+\ln(1+1/(N−1)^2)+⋯+\ln(1+1/4).\) Tangu\(\ln(1+t)\) imepakana na mara kwa mara\(t\), wakati\(0<t<1\) mtu ana\(\displaystyle \ln(S_N)≤C\sum_{n=1}^N\frac{1}{n^2}\), ambayo hujiunga kwa kulinganisha na\(p\) -mfululizo kwa\(p=2\).