Loading [MathJax]/extensions/TeX/boldsymbol.js
Skip to main content
Library homepage
 
Global

9.4: Vipimo vya kulinganisha

  • Edwin “Jed” Herman & Gilbert Strang
  • OpenStax

Malengo ya kujifunza
  • 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.

Hii inaonyesha grafu mbili kwa upande. Ya kwanza inaonyesha pointi zilizopangwa kwa kiasi cha sehemu kwa jumla ya 1/n ^ 2 na jumla 1/ (n ^ 2 + 1). Kila moja ya kiasi cha sehemu ya mwisho ni chini ya jumla ya sambamba ya sehemu ya zamani. Ya pili inaonyesha pointi zilizopangwa kwa kiasi cha sehemu kwa jumla ya 1/ (n - 0.5) na jumla 1/n.Kila moja ya kiasi cha sehemu ya mwisho ni chini ya jumla ya sehemu ya sambamba ya zamani.
Kielelezo\PageIndex{1}: (a) Kila moja ya kiasi cha sehemu ya mfululizo uliopewa ni chini ya sambamba sehemu ya jumla kwa convergingp−series. (b) Kila moja ya kiasi sehemu ya mfululizo kutokana ni kubwa kuliko sambamba sehemu jumla kwa diverging harmonic mfululizo.
Jedwali\PageIndex{1}: Kulinganisha mfululizo nap -mfululizo (p = 2)
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
Jedwali\PageIndex{2}: Kulinganisha mfululizo na mfululizo wa harmonic
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
Mtihani wa kulinganisha
  1. 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.
  2. 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.
Ushahidi

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.

Mfano\PageIndex{1}: Using the Comparison Test

Kwa kila mfululizo wafuatayo, tumia mtihani wa kulinganisha ili uone kama mfululizo unajiunga au hupungua.

  1. \displaystyle \sum_{n=1}^∞=\dfrac{1}{n^3+3n+1}
  2. \displaystyle \sum_{n=1}^∞=\dfrac{1}{2^n+1}
  3. \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.

Zoezi\PageIndex{1}

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.

Limit kulinganisha mtihani

Hebua_n,b_n≥0 kwa woten≥1.

  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.
  2. 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.
  3. 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.

Mfano\PageIndex{2}: Using the Limit Comparison Test

Kwa kila moja ya mfululizo wafuatayo, tumia mtihani wa kulinganisha kikomo ili uone kama mfululizo unajiunga au hupungua. Ikiwa mtihani hautumiki, sema hivyo.

  1. \displaystyle \sum^∞_{n=1}\dfrac{1}{\sqrt{n}+1}
  2. \displaystyle \sum^∞_{n=1}\dfrac{2^n+1}{3^n}
  3. \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.

Zoezi\PageIndex{2}

Tumia mtihani wa kulinganisha kikomo ili uone kama mfululizo\displaystyle \sum^∞_{n=1}\dfrac{5^n}{3^n+2} unajiunga au hupungua.

Kidokezo

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.