9.4: Vipimo vya kulinganisha
- Tumia mtihani wa kulinganisha ili kupima mfululizo wa kuunganisha.
- Tumia mtihani wa kulinganisha kikomo ili kuamua muunganiko wa mfululizo.
Tumeona kwamba mtihani muhimu inaruhusu sisi kuamua muunganiko au tofauti ya mfululizo kwa kulinganisha na muhimu kuhusiana yasiyofaa. Katika sehemu hii, tunaonyesha jinsi ya kutumia vipimo vya kulinganisha ili kuamua muunganiko au tofauti ya mfululizo kwa kulinganisha na mfululizo ambao muunganiko au tofauti yake hujulikana. Kwa kawaida vipimo hivi hutumiwa kuamua muunganiko wa mfululizo ambao ni sawa na mfululizo wa kijiometri aup -mfululizo.
Mtihani wa kulinganisha
Katika sehemu mbili zilizopita, tulijadili madarasa mawili makubwa ya mfululizo: mfululizo wa kijiometri nap -mfululizo. Tunajua hasa wakati mfululizo huu hujiunga na wakati wanapotofautiana. Hapa tunaonyesha jinsi ya kutumia muunganiko au upatanisho wa mfululizo huu ili kuthibitisha muunganiko au tofauti kwa mfululizo mwingine, kwa kutumia njia inayoitwa mtihani wa kulinganisha.
Kwa mfano, fikiria mfululizo
\sum_{n=1}^∞\dfrac{1}{n^2+1}. \nonumber
Mfululizo huu unaonekana sawa na mfululizo unaobadilika
\sum_{n=1}^∞\dfrac{1}{n^2} \nonumber
Kwa kuwa maneno katika kila mfululizo ni chanya, mlolongo wa kiasi cha sehemu kwa kila mfululizo ni kuongezeka kwa monotone. Aidha, tangu
0<\dfrac{1}{n^2+1}<\dfrac{1}{n^2} \nonumber
kwa integers wote chanyan, jumla yak^{\text{th}} sehemuS_k ya\displaystyle \sum^∞_{n=1}\dfrac{1}{n^2+1} satisfies
S_k=\sum_{n=1}^k\dfrac{1}{n^2+1}<\sum_{n=1}^k\dfrac{1}{n^2}<\sum_{n=1}^∞\dfrac{1}{n^2}. \nonumber
(Angalia Kielelezo\PageIndex{1a} na Jedwali\PageIndex{1}.) Tangu mfululizo juu ya haki hujiunga, mlolongo{S_k} umefungwa hapo juu. Tunahitimisha kwamba{S_k} ni monotone kuongeza mlolongo kwamba ni imepakana juu. Kwa hiyo, kwa Theorem Monotone Convergence,{S_k} converges, na hivyo
\sum_{n=1}^∞\dfrac{1}{n^2+1} \nonumber
hukutana.
Vile vile, fikiria mfululizo
\sum_{n=1}^∞\dfrac{1}{n−1/2}. \nonumber
Mfululizo huu unaonekana sawa na mfululizo tofauti
\sum_{n=1}^∞\dfrac{1}{n}. \nonumber
Mlolongo wa kiasi cha sehemu kwa kila mfululizo ni monotone kuongezeka na
\dfrac{1}{n−1/2}>\dfrac{1}{n}>0 \nonumber
kwa kila integer chanyan. Kwa hiyo, jumla yak^{\text{th}} sehemuS_k ya
\sum^∞_{n=1}\dfrac{1}{n−1/2} \nonumber
hushibisha
S_k=\sum_{n=1}^k\dfrac{1}{n−1/2}>\sum_{n=1}^k\dfrac{1}{n}. \nonumber
(Angalia Kielelezo\PageIndex{1n} na Jedwali\PageIndex{1}). Kwa kuwa mfululizo\displaystyle \sum^∞_{n=1}\frac{1}{n} unajitokeza kwa infinity, mlolongo wa kiasi cha sehemu\displaystyle \sum^k_{n=1}\frac{1}{n} ni unbounded. Kwa hiyo,{S_k} ni mlolongo unbounded, na kwa hiyo diverges. Tunahitimisha kwamba
\sum_{n=1}^∞\dfrac{1}{n−1/2} \nonumber
hutengana.

k | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
---|---|---|---|---|---|---|---|---|
\displaystyle \sum_{n=1}^k\dfrac{1}{n^2+1} | 0.5 | 0.7 | 0.8 | 0.8588 | 0.8973 | 0.9243 | 0.9443 | 0.9597 |
\displaystyle \sum_{n=1}^k\dfrac{1}{n^2} | 1 | 1.25 | 1.3611 | 1.4236 | 1.4636 | 1.4914 | 1.5118 | 1.5274 |
k | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
---|---|---|---|---|---|---|---|---|
\displaystyle \sum_{n=1}^k\dfrac{1}{n−1/2} | 2 | 2.6667 | 3.0667 | 3.3524 | 3.5746 | 3.7564 | 3.9103 | 4.0436 |
\displaystyle \sum_{n=1}^k\dfrac{1}{n} | 1 | 1.5 | 1.8333 | 2.0933 | 2.2833 | 2.45 | 2.5929 | 2.7179 |
- Tuseme kuna integerN such that 0≤a_n≤b_n for all n≥N. If \displaystyle \sum^∞_{n=1}b_n converges, then \displaystyle \sum^∞_{n=1}a_n converges.
- Tuseme kuna integerN such that a_n≥b_n≥0 for all n≥N. If \displaystyle \sum^∞_{n=1}b_n diverges, then \displaystyle \sum^∞_{n=1}a_n diverges.
Sisi kuthibitisha sehemu i. ushahidi wa sehemu ya ii. ni contrapositive ya sehemu i. basi{S_k} be the sequence of partial sums associated with \displaystyle \sum^∞_{n=1}a_n, and let \displaystyle L=\sum^∞_{n=1}b_n. Since the terms a_n≥0,
S_k=a_1+a_2+⋯+a_k≤a_1+a_2+⋯+a_k+a_{k+1}=S_{k+1}. \nonumber
Kwa hiyo, mlolongo wa kiasi cha sehemu huongezeka. Zaidi ya hayo, tangua_n≤b_n for all n≥N, then
\sum_{n=N}^ka_n≤\sum_{n=N}^kb_n≤\sum_{n=1}^∞b_n=L. \nonumber
Kwa hiyo, kwa wotek≥1,
S_k=(a_1+a_2+⋯+a_{N−1})+\sum_{n=N}^ka_n≤(a_1+a_2+⋯+a_{N−1})+L. \nonumber
Tangua_1+a_2+⋯+a_{N−1} is a finite number, we conclude that the sequence {S_k} is bounded above. Therefore, {S_k} is an increasing sequence that is bounded above. By the Monotone Convergence Theorem, we conclude that {S_k} converges, and therefore the series \displaystyle \sum_{n=1}^∞a_n converges.
□
Kutumia mtihani wa kulinganisha ili kuamua muunganiko au tofauti ya mfululizo\displaystyle \sum_{n=1}^∞a_n, it is necessary to find a suitable series with which to compare it. Since we know the convergence properties of geometric series and p-series, these series are often used. If there exists an integer N such that for all n≥N, each term an is less than each corresponding term of a known convergent series, then \displaystyle \sum_{n=1}^∞a_n converges. Similarly, if there exists an integer N such that for all n≥N, each term an is greater than each corresponding term of a known divergent series, then \displaystyle \sum_{n=1}^∞a_n diverges.
Kwa kila mfululizo wafuatayo, tumia mtihani wa kulinganisha ili uone kama mfululizo unajiunga au hupungua.
- \displaystyle \sum_{n=1}^∞=\dfrac{1}{n^3+3n+1}
- \displaystyle \sum_{n=1}^∞=\dfrac{1}{2^n+1}
- \displaystyle \sum_{n=2}^∞=\dfrac{1}{\ln \,n }
Suluhisho
a. kulinganisha na\displaystyle \sum_{n=1}^∞\dfrac{1}{n^3}. Kwa kuwa\displaystyle \sum_{n=1}^∞\dfrac{1}{n^3} nip -mfululizo nap=3, hujiunga. Zaidi ya hayo,
\dfrac{1}{n^3+3n+1}<\dfrac{1}{n^3} \nonumber
kwa kila integer chanyan. Kwa hiyo, tunaweza kuhitimisha kwamba\displaystyle \sum^∞_{n=1}\dfrac{1}{n^3+3n+1} hujiunga.
b. kulinganisha na\displaystyle \sum^∞_{n=1}\left(\dfrac{1}{2}\right)^n. Kwa kuwa\displaystyle \sum_{n=1}^∞\left(\dfrac{1}{2}\right)^n ni mfululizo wa kijiometrir=\dfrac{1}{2} na na\left|\dfrac{1}{2}\right|<1, hujiunga. Pia,
\dfrac{1}{2^n+1}<\dfrac{1}{2^n} \nonumber
kwa kila integer chanyan. Kwa hiyo, tunaona kwamba\displaystyle \sum^∞_{n=1}\dfrac{1}{2^n+1} hujiunga.
c. kulinganisha na\displaystyle \sum^∞_{n=2}\dfrac{1}{n}. Tangu
\dfrac{1}{\ln n }>\dfrac{1}{n} \nonumber
kwa kila integern≥2 na\displaystyle \sum^∞_{n=2}\dfrac{1}{n} diverges, tuna kwamba\displaystyle \sum^∞_{n=2}\dfrac{1}{\ln n} diverges.
Tumia mtihani wa kulinganisha ili uone ikiwa mfululizo\displaystyle \sum^∞_{n=1}\dfrac{n}{n^3+n+1} unajiunga au unapungua.
- Kidokezo
-
Pata thamanip kama hiyo\dfrac{n}{n^3+n+1}≤\dfrac{1}{n^p}.
- Jibu
-
Mfululizo hujiunga.
Limit kulinganisha mtihani
mtihani kulinganisha kazi vizuri kama tunaweza kupata mfululizo kulinganishwa kuridhisha hypothesis ya mtihani. Hata hivyo, wakati mwingine kutafuta mfululizo sahihi inaweza kuwa vigumu. Fikiria mfululizo
\sum_{n=2}^∞\dfrac{1}{n^2−1}. \nonumber
Ni kawaida kulinganisha mfululizo huu na mfululizo unaobadilika.
\sum_{n=2}^∞\dfrac{1}{n^2}. \nonumber
Hata hivyo, mfululizo huu haikidhi hypothesis muhimu kutumia mtihani wa kulinganisha kwa sababu
\dfrac{1}{n^2−1}>\dfrac{1}{n^2} \nonumber
kwa integers woten≥2. Ingawa tunaweza kuangalia kwa mfululizo tofauti na ambayo kulinganisha\displaystyle \sum^∞_{n=2}\frac{1}{n^2−1}, badala sisi kuonyesha jinsi tunaweza kutumia kikomo kulinganisha mtihani kulinganisha
\sum_{n=2}^∞\frac{1}{n^2−1} \nonumber
na
\sum_{n=2}^∞\frac{1}{n^2}. \nonumber
Hebu tuchunguze wazo nyuma ya mtihani wa kulinganisha kikomo. Fikiria mbili mfululizo\displaystyle \sum^∞_{n=1}a_n\displaystyle \sum^∞_{n=1}b_n na. kwa maneno mazuria_nb_n na kutathmini
\lim_{n→∞}\frac{a_n}{b_n}. \nonumber
Kama
\lim_{n→∞}\frac{a_n}{b_n}=L≠0, \nonumber
basi, kwan kutosha kubwa,a_n≈Lb_n. Kwa hiyo, ama mfululizo wote hujiunga au mfululizo wote hutofautiana. Kwa mfululizo\displaystyle \sum^∞_{n=2}\frac{1}{n^2−1} na\displaystyle \sum^∞_{n=2}\dfrac{1}{n^2}, tunaona kwamba
\lim_{n→∞}\dfrac{1/(n^2−1)}{1/n^2}=\lim_{n→∞}\dfrac{n^2}{n^2−1}=1. \nonumber
Tangu\displaystyle \sum^∞_{n=2}\frac{1}{n^2} converges, sisi kuhitimisha kwamba
\sum_{n=2}^∞\dfrac{1}{n^2−1} \nonumber
hukutana.
Mtihani wa kulinganisha kikomo unaweza kutumika katika kesi nyingine mbili. Tuseme
\lim_{n→∞}\dfrac{a_n}{b_n}=0. \nonumber
Katika kesi hii,{a_n/b_n} ni mlolongo imepakana. Matokeo yake, kuna mara kwa maraM kama hiyoa_n≤Mb_n. Kwa hiyo, ikiwa\displaystyle \sum^∞_{n=1}b_n hujiunga, kisha\displaystyle \sum^∞_{n=1}a_n hujiunga. Kwa upande mwingine, tuseme
\lim_{n→∞}\dfrac{a_n}{b_n}=∞. \nonumber
Katika kesi hii,{a_n/b_n} ni mlolongo unbounded. Kwa hiyo, kwa kila maraM kuna integerN kama hiyoa_n≥Mb_n kwa woten≥N. Kwa hiyo,\displaystyle \sum^∞_{n=1}b_n ikiwa hupungua,\displaystyle \sum^∞_{n=1}a_n kisha hupungua pia.
Hebua_n,b_n≥0 kwa woten≥1.
- Ikiwa\displaystyle \lim_{n→∞}\frac{a_n}{b_n}=L≠0, basi\displaystyle \sum^∞_{n=1}a_n na\displaystyle \sum^∞_{n=1}b_n wote wawili hujiunga au wote wawili wanatofautiana.
- Ikiwa\displaystyle \lim_{n→∞}\frac{a_n}{b_n}=0 na\displaystyle \sum^∞_{n=1}b_n hujiunga, kisha\displaystyle \sum^∞_{n=1}a_n hujiunga.
- Ikiwa\displaystyle \lim_{n→∞}\frac{a_n}{b_n}=∞ na\displaystyle \sum^∞_{n=1}b_n hutofautiana, basi\displaystyle \sum^∞_{n=1}a_n hupungua.
Kumbuka kwamba ikiwa\dfrac{a_n}{b_n}→0 na\displaystyle \sum^∞_{n=1}b_n hutofautiana, mtihani wa kulinganisha kikomo hautoi habari. Vile vile, ikiwa\dfrac{a_n}{b_n}→∞ na\displaystyle \sum^∞_{n=1}b_n hujiunga, mtihani pia hautoi habari. Kwa mfano, fikiria mfululizo mbili\displaystyle \sum_{n=1}^∞\frac{1}{\sqrt{n}} na\displaystyle \sum_{n=1}^∞\frac{1}{n^2}. Mfululizo huu ni wotep -mfululizop=\frac{1}{2} nap=2, kwa mtiririko huo. Tangup=\frac{1}{2}<1, mfululizo\displaystyle \sum_{n=1}^∞\frac{1}{\sqrt{n}} hupungua. Kwa upande mwingine, tangup=2>1, mfululizo\displaystyle \sum_{n=1}^∞\frac{1}{n^2} hujiunga. Hata hivyo, tuseme tulijaribu kutumia mtihani wa kulinganisha kikomo, kwa kutumia pmfululizo wa mfululizo\displaystyle \sum_{n=1}^∞\frac{1}{n^3} kama mfululizo wetu wa kulinganisha. Kwanza, tunaona kwamba
\dfrac{1/\sqrt{n}}{1/n^3}=\dfrac{n^3}{\sqrt{n}}=n^{5/2}→∞\; \text{ as } \;n→∞. \nonumber
Vile vile, tunaona kwamba
\dfrac{1/n^2}{1/n^3}=n→∞\; \text{ as } \;n→∞. \nonumber
Kwa hiyo, kama\dfrac{a_n}{b_n}→∞ wakati\displaystyle \sum_{n=1}^∞b_n converges, hatuwezi kupata taarifa yoyote juu ya muunganiko au tofauti ya\displaystyle \sum_{n=1}^∞a_n.
Kwa kila moja ya mfululizo wafuatayo, tumia mtihani wa kulinganisha kikomo ili uone kama mfululizo unajiunga au hupungua. Ikiwa mtihani hautumiki, sema hivyo.
- \displaystyle \sum^∞_{n=1}\dfrac{1}{\sqrt{n}+1}
- \displaystyle \sum^∞_{n=1}\dfrac{2^n+1}{3^n}
- \displaystyle \sum^∞_{n=1}\dfrac{\ln(n)}{n^2}
Suluhisho
a. kulinganisha mfululizo huu kwa\displaystyle \sum^∞_{n=1}\dfrac{1}{\sqrt{n}}. Tumia
\displaystyle \lim_{n→∞}\dfrac{1/(\sqrt{n}+1)}{1/\sqrt{n}}=\lim_{n→∞}\dfrac{\sqrt{n}}{\sqrt{n}+1}=\lim_{n→∞}\dfrac{1/\sqrt{n}}{1+1/\sqrt{n}}=1.
Kwa mtihani wa kulinganisha kikomo,\displaystyle \sum^∞_{n=1}\dfrac{1}{\sqrt{n}} tangu hupungua, kisha\displaystyle \sum^∞_{n=1}\dfrac{1}{\sqrt{n}+1} hupungua.
b. kulinganisha mfululizo huu kwa\displaystyle \sum^∞_{n=1}\left(\dfrac{2}{3}\right)^n. Tunaona kwamba
\displaystyle \lim_{n→∞}\dfrac{(2^n+1)/3^n}{2^n/3^n}=\lim_{n→∞}\dfrac{2^n+1}{3^n}⋅\dfrac{3^n}{2^n}=\lim_{n→∞}\dfrac{2^n+1}{2^n}=\lim_{n→∞}\left[1+\left(\tfrac{1}{2}\right)^n\right]=1.
Kwa hiyo,
\displaystyle \lim_{n→∞}\dfrac{(2^n+1)/3^n}{2^n/3^n}=1.
Tangu\displaystyle \sum^∞_{n=1}\left(\dfrac{2}{3}\right)^n hujiunga, tunahitimisha kuwa\displaystyle \sum^∞_{n=1}\dfrac{2^n+1}{3^n} hujiunga.
c. tangu\ln n<n, kulinganisha na\displaystyle \sum^∞_{n=1}\dfrac{1}{n}. Tunaona kwamba
\displaystyle \lim_{n→∞}\dfrac{\ln n/n^2}{1/n}=\lim_{n→∞}\dfrac{\ln n}{n^2}⋅\dfrac{n}{1}=\lim_{n→∞}\dfrac{\ln n}{n}.
Ili kutathmini\displaystyle \lim_{n→∞}\ln n/n, tathmini kikomo kamax→∞ ya kazi halisi ya thamani\ln(x)/x. Mipaka hii miwili ni sawa, na kufanya mabadiliko haya inatuwezesha kutumia utawala wa L'Hôpital. Tunapata
\displaystyle \lim_{x→∞}\dfrac{lnx}{x}=\lim_{x→∞}\dfrac{1}{x}=0.
Kwa hiyo\displaystyle \lim_{n→∞}\frac{\ln n}{n}=0, na, kwa hiyo,
\displaystyle \lim_{n→∞}\dfrac{(\ln n)/n^2}{1/n}=0.
Kwa kuwa kikomo ni0 lakini\displaystyle \sum^∞_{n=1}\dfrac{1}{n} kinapungua, mtihani wa kulinganisha kikomo haitoi taarifa yoyote.
Linganisha na\displaystyle \sum^∞_{n=1}\dfrac{1}{n^2} badala yake. Katika kesi hiyo,
\displaystyle \lim_{n→∞}\dfrac{(\ln n)/n^2}{1/n^2}=\lim_{n→∞}\dfrac{\ln n}{n^2}⋅\dfrac{n^2}{1}=\lim_{n→∞}\ln n=∞.
Kwa kuwa kikomo ni∞ lakini\displaystyle \sum^∞_{n=1}\dfrac{1}{n^2} hujiunga, mtihani bado hautoi taarifa yoyote.
Kwa hiyo sasa tunajaribu mfululizo kati ya mbili ambazo tayari tumejaribu. Kuchagua mfululizo\displaystyle \sum^∞_{n=1}\dfrac{1}{n^{3/2}}, tunaona kwamba
\displaystyle \lim_{n→∞}\dfrac{(\ln n)/n^2}{1/n^{3/2}}=\lim_{n→∞}\dfrac{\ln n}{n^2}⋅\dfrac{n^{3/2}}{1}=\lim_{n→∞}\dfrac{\ln n}{\sqrt{n}}.
Kama hapo juu, ili kutathmini\displaystyle \lim_{n→∞}\frac{\ln n}{\sqrt{n}}, tathmini kikomo kamax→∞ ya kazi halisi ya thamani\frac{\ln n}{\sqrt{n}}. Kwa kutumia utawala wa L'Hôpital,
\displaystyle \lim_{x→∞}\dfrac{\ln x}{\sqrt{x}}=\lim_{x→∞}\dfrac{2\sqrt{x}}{x}=\lim_{x→∞}\dfrac{2}{\sqrt{x}}=0.
Kwa kuwa kikomo ni0 na\displaystyle \sum^∞_{n=1}\dfrac{1}{n^{3/2}} hujiunga, tunaweza kuhitimisha kwamba\displaystyle \sum^∞_{n=1}\dfrac{\ln n}{n^2} hujiunga.
Tumia mtihani wa kulinganisha kikomo ili uone kama mfululizo\displaystyle \sum^∞_{n=1}\dfrac{5^n}{3^n+2} unajiunga au hupungua.
- Kidokezo
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Linganisha na mfululizo wa kijiometri.
- Jibu
-
Mfululizo hutofautiana.
Dhana muhimu
- Vipimo vya kulinganisha hutumiwa kuamua kuunganishwa au kutofautiana kwa mfululizo na maneno mazuri.
- Wakati wa kutumia vipimo vya kulinganisha, mfululizo mara nyingi\displaystyle \sum^∞_{n=1}a_n hulinganishwa na kijiometri aup -mfululizo.
faharasa
- mtihani wa kulinganisha
- Ikiwa0≤a_n≤b_n kwa woten≥N na\displaystyle \sum^∞_{n=1}b_n hujiunga, basi\displaystyle \sum^∞_{n=1}a_n hujiunga; ikiwaa_n≥b_n≥0 kwa woten≥N\displaystyle \sum^∞_{n=1}b_n na hupungua, basi\displaystyle \sum^∞_{n=1}a_n hupungua.
- mtihani wa kulinganisha kikomo
- Tusemea_n,b_n≥0 kwa woten≥1. Ikiwa\displaystyle \lim_{n→∞}a_n/b_n→L≠0, basi\displaystyle \sum^∞_{n=1}a_n na\displaystyle \sum^∞_{n=1}b_n wote wawili hujiunga au wote wawili wanatofautiana; ikiwa\displaystyle \lim_{n→∞}a_n/b_n→0 na\displaystyle \sum^∞_{n=1}b_n hujiunga, kisha\displaystyle \sum^∞_{n=1}a_n hujiunga. Ikiwa\displaystyle \lim_{n→∞}a_n/b_n→∞, na\displaystyle \sum^∞_{n=1}b_n hupungua, basi\displaystyle \sum^∞_{n=1}a_n hupungua.