10.10-bo'lim: Nisbat sinovi
Ushbu bo'limda biz seriyaning mutlaq yaqinlashishi yoki yo'qligini aniqlash uchun foydalanishimiz mumkin bo'lgan testni ko'rib chiqamiz. Eslatib o'tamiz, agar qator mutlaqo yaqin bo'lsa, biz uning konvergent ekanligini ham bilib olamiz va shuning uchun biz ko'pincha ketma-ketlikning yaqinlashuvini aniqlash uchun foydalanamiz.
Sinovni davom ettirishdan oldin, faktoriallarni tezda eslatib o'tamiz. Ushbu test, ayniqsa, faktoriallarni o'z ichiga olgan seriyalar uchun foydali bo'ladi (va biz ba'zilarini ilovalarda ko'ramiz), shuning uchun ularni misolda ko'rishdan oldin ularni hal qilishimizga ishonch hosil qilaylik.
Agar \(n\) \(n \ge 0\) ga teng butun son bo'lsa, \(n\) faktorial quyidagicha aniqlanadi:
\[\begin{align*}n! & = n\left( {n - 1} \right)\left( {n - 2} \right) \cdots \left( 3 \right)\left( 2 \right)\left( 1 \right) & \hspace{0.15in} & {\mbox{if }}n \ge 1\\ 0! & = 1 & \hspace{0.15in} & {\mbox{by definition}}\end{align*}\]Keling, bir juftlikni tezda hisoblaylik.
\[\begin{align*}& 1! = 1\\ & 2! = 2\left( 1 \right) = 2\\ & 3! = 3\left( 2 \right)\left( 1 \right) = 6\\ & 4! = 4\left( 3 \right)\left( 2 \right)\left( 1 \right) = 24\\ & 5! = 5\left( 4 \right)\left( 3 \right)\left( 2 \right)\left( 1 \right) = 120\end{align*}\]Yuqoridagi oxirgi hisob-kitobda biz faktorialni bir necha xil usulda qayta yozishimiz mumkinligiga e'tibor bering. Masalan,
\[\begin{align*}5! & = 5\underbrace {\left( 4 \right)\left( 3 \right)\left( 2 \right)\left( 1 \right)}_{4!} = 5 \cdot 4!\\ 5! & = 5\left( 4 \right)\underbrace {\left( 3 \right)\left( 2 \right)\left( 1 \right)}_{3!} = 5\left( 4 \right) \cdot 3!\end{align*}\]Umuman olganda, biz har doim faktorialdan atamalarni quyidagicha “chiqib tashlashimiz” mumkin.
\[\begin{align*}n! & = n\left( {n - 1} \right)\left( {n - 2} \right) \cdots \left( {n - k} \right)\left( {n - \left( {k + 1} \right)} \right) \cdots \left( 3 \right)\left( 2 \right)\left( 1 \right)\\ & = n\left( {n - 1} \right)\left( {n - 2} \right) \cdots \left( {n - k} \right) \cdot \left( {n - \left( {k + 1} \right)} \right)!\\ & = n\left( {n - 1} \right)\left( {n - 2} \right) \cdots \left( {n - k} \right) \cdot \left( {n - k - 1} \right)!\end{align*}\]Biz buni vaqti-vaqti bilan qilishimiz kerak, shuning uchun bu haqda unutmang.
Bundan tashqari, faktoriallar bilan ishlashda biz qavslar bilan juda ehtiyot bo'lishimiz kerak. Masalan, \(\left( {2n} \right)! \ne 2\,\,n!\), agar biz quyidagi faktoriallarning har birini yozsak, ko'ramiz.
\[\begin{align*}\left( {2n} \right)! & = \left( {2n} \right)\left( {2n - 1} \right)\left( {2n - 2} \right) \cdots \left( 3 \right)\left( 2 \right)\left( 1 \right)\\ 2\,\,n! & = 2\left[ {\left( n \right)\left( {n - 1} \right)\left( {n - 2} \right) \cdots \left( 3 \right)\left( 2 \right)\left( 1 \right)} \right]\end{align*}\]Shunga qaramay, biz qavslar bilan faktoriallarni ko'rib chiqamiz, shuning uchun ularni tashlamang. Bu ko'pincha o'quvchilar faktoriallar bo'ylab birinchi marta yugurganlarida tez-tez uchraydigan xatolardan biridir.
OK, biz hozir testga tayyormiz.
Nisbat sinovi
Bizda \(\displaystyle \sum {{a_n}} \) seriyasi bor deylik. Aniqlash,
\[L = \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{a_{n + 1}}}}{{{a_n}}}} \right|\]Keyin,
- agar \(L < 1\) seriyasi mutlaqo yaqinlashuvchi (va shuning uchun konvergent).
- agar \(L > 1\) seriya farqlanadi.
- agar \(L = 1\) qator divergent, shartli konvergent yoki mutlaqo yaqinlashuvchi bo'lishi mumkin.
A dalil Ushbu test bo'limning oxirida.
E'tibor bering, \(L = 1\) holatida nisbat testi deyarli foydasiz va biz ketma-ketlikning yaqinligini aniqlash uchun boshqa testga murojaat qilishimiz kerak.
Bundan tashqari, \(L\) ta'rifidagi mutlaq qiymat satrlari mutlaqo talab qilinadi. Agar ular bo'lmasa, biz to'g'ri javob olishimiz mumkin emas.
Keling, ba'zi misollarni ko'rib chiqaylik.
Yechim
Ushbu birinchi misolda keling, biroz ehtiyot bo'laylik va hamma narsa to'g'ri yozilganligiga ishonch hosil qilaylik. Mana \({a_n}\) seriyali atamalar.
\[{a_n} = \frac{{{{\left( { - 10} \right)}^n}}}{{{4^{2n + 1}}\left( {n + 1} \right)}}\]Eslatib o'tamiz, \({a_{n + 1}}\) ni hisoblash uchun faqat almashtirish kerak n+1 \({a_n}\)-dagi barcha \(n\) lar uchun.
\[{a_{n + 1}} = \frac{{{{\left( { - 10} \right)}^{n + 1}}}}{{{4^{2\left( {n + 1} \right) + 1}}\left( {\left( {n + 1} \right) + 1} \right)}} = \frac{{{{\left( { - 10} \right)}^{n + 1}}}}{{{4^{2n + 3}}\left( {n + 2} \right)}}\]Endi \(L\) ni aniqlash uchun biz foydalanamiz,
\[L = \mathop {\lim }\limits_{n \to \infty } \left| {{a_{n + 1}} \cdot \frac{1}{{{a_n}}}} \right|\]chunki kasrlar bilan ishlashda bu bizda biroz osonroq bo'ladi. Shunday qilib,
\[\begin{align*}L & = \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{{\left( { - 10} \right)}^{n + 1}}}}{{{4^{2n + 3}}\left( {n + 2} \right)}}\,\,\frac{{{4^{2n + 1}}\left( {n + 1} \right)}}{{{{\left( { - 10} \right)}^n}}}} \right|\\ & = \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{ - 10\left( {n + 1} \right)}}{{{4^2}\left( {n + 2} \right)}}} \right|\\ & = \frac{{10}}{{16}}\mathop {\lim }\limits_{n \to \infty } \frac{{n + 1}}{{n + 2}}\\ & = \frac{{10}}{{16}} < 1\end{align*}\]Shunday qilib, \(L < 1\) va shunga o'xshash nisbatlar testi bo'yicha qatorlar mutlaqo yaqinlashadi va shuning uchun yaqinlashadi.
Oldingi misolda ko'rinib turganidek, ularda odatda juda ko'p bekor qilishlar sodir bo'ladi. Buni bekor qilganingizga ishonch hosil qiling. Agar siz bunday bekor qilmasangiz, bu cheklovni ancha qiyinlashtirishi mumkin.
Yechim
Endi biz batafsil ishlaganimizdan so'ng, qolganlari bilan to'liq tafsilotga kirmaymiz. Mana chegara.
\[L = \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{\left( {n + 1} \right)!}}{{{5^{n + 1}}}}\,\frac{{{5^n}}}{{n!}}} \right| = \mathop {\lim }\limits_{n \to \infty } \frac{{\left( {n + 1} \right)!}}{{5\,\,n!}}\]Ushbu cheklovni bajarish uchun biz faktorlarni yo'q qilishimiz kerak. Biz undagi faktoriallar bilan cheklovni bajara olmaymiz. Faktoriallarni yo'q qilish uchun biz faktoriallar haqidagi yuqoridagi munozarada eslaymizki, biz har doim faktorialdan atamalarni "olib tashlashimiz" mumkin. Agar biz buni surat bilan qilsak (bu holda u ikkitadan kattaroq bo'lgani uchun) biz olamiz,
\[L = \mathop {\lim }\limits_{n \to \infty } \frac{{\left( {n + 1} \right)\,\,n!}}{{5\,\,n!}}\]qaysi nuqtada biz \(n\) ni bekor qilishimiz mumkin! hisoblagich olish uchun maxraj,
\[L = \mathop {\lim }\limits_{n \to \infty } \frac{{\left( {n + 1} \right)}}{5} = \infty > 1\]Shunday qilib, nisbat testi bo'yicha bu seriya farqlanadi.
Yechim
Bunday holda, faktorlar bilan ishlashda ehtiyot bo'ling.
\[\begin{align*}L & = \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{{\left( {n + 1} \right)}^2}}}{{\left( {2\left( {n + 1} \right) - 1} \right)!}}\,\frac{{\left( {2n - 1} \right)!}}{{{n^2}}}} \right|\\ & = \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{{\left( {n + 1} \right)}^2}}}{{\left( {2n + 1} \right)!}}\,\frac{{\left( {2n - 1} \right)!}}{{{n^2}}}} \right|\\ & = \mathop {\lim }\limits_{n \to \infty } \frac{{{{\left( {n + 1} \right)}^2}}}{{\left( {2n + 1} \right)\left( {2n} \right)\left( {2n - 1} \right)!}}\,\frac{{\left( {2n - 1} \right)!}}{{{n^2}}}\\ & = \mathop {\lim }\limits_{n \to \infty } \frac{{{{\left( {n + 1} \right)}^2}}}{{\left( {2n + 1} \right)\left( {2n} \right)\left( {{n^2}} \right)}}\\ & = 0 < 1\end{align*}\]Shunday qilib, nisbat testi bo'yicha bu qator mutlaqo yaqinlashadi va shuning uchun yaqinlashadi.
Yechim
Buni a deb xato qilmang geometrik qator. Maxrajdagi \(n\) bu geometrik qator emasligini bildiradi. Shunday qilib, chegarani hisoblaylik.
\[\begin{align*}L & = \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{9^{n + 1}}}}{{{{\left( { - 2} \right)}^{n + 2}}\,\left( {n + 1} \right)}}\,\,\frac{{{{\left( { - 2} \right)}^{n + 1}}\,\,n}}{{{9^n}}}} \right|\\ & = \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{9\,n}}{{\left( { - 2} \right)\,\left( {n + 1} \right)}}} \right|\\ & = \frac{9}{2}\mathop {\lim }\limits_{n \to \infty } \frac{n}{{n + 1}}\\ & = \frac{9}{2} > 1\end{align*}\]Shuning uchun, nisbat testiga ko'ra, bu seriya farqlanadi.
Oldingi misolda to'g'ri javob olish uchun mutlaq qiymat chiziqlari talab qilingan. Agar biz ulardan foydalanmaganimizda, biz \(L = - \frac{9}{2} < 1\)-ni olgan bo'lardik, bu konvergent qatorni nazarda tutgan bo'lar edi!
Keling, \(L = 1\) ni olganimizda nima sodir bo'lishini bilish uchun bir nechta misollarni ko'rib chiqaylik. Eslatib o'tamiz, nisbat testi bizga ushbu seriyalarning yaqinlashuvi haqida hech narsa aytmaydi. Ushbu ikkala misolda biz avval \(L = 1\) olganimizni tekshiramiz va keyin konvergentsiyani aniqlash uchun boshqa testlardan foydalanamiz.
Yechim
Keling, avval \(L\) ni olamiz.
\[L = \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{{\left( { - 1} \right)}^{n + 1}}}}{{{{\left( {n + 1} \right)}^2} + 1}}\,\,\frac{{{n^2} + 1}}{{{{\left( { - 1} \right)}^n}}}} \right| = \mathop {\lim }\limits_{n \to \infty } \frac{{{n^2} + 1}}{{{{\left( {n + 1} \right)}^2} + 1}} = 1\]Shunday qilib, yuqorida aytib o'tilganidek, biz \(L = 1\) ni olamiz, ya'ni nisbat testi ushbu seriyaning yaqinlashuvini aniqlash uchun yaxshi emas. Biz ushbu seriya uchun boshqa sinovga murojaat qilishimiz kerak. Ushbu seriya o'zgaruvchan seriyadir va shuning uchun keling, ushbu testdagi ikkita shartni tekshirib ko'raylik.
\[\mathop {\lim }\limits_{n \to \infty } {b_n} = \mathop {\lim }\limits_{n \to \infty } \frac{1}{{{n^2} + 1}} = 0\] \[{b_n} = \frac{1}{{{n^2} + 1}} > \frac{1}{{{{\left( {n + 1} \right)}^2} + 1}} = {b_{n + 1}}\]Ikki shart bajariladi va shuning uchun muqobil seriyali test orqali bu seriya konvergent hisoblanadi. Bu seriyaning ham mutlaqo konvergentligini tekshirishni sizga qoldiramiz.
Yechim
Mana chegara.
\[L = \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{n + 3}}{{2\left( {n + 1} \right) + 7}}\,\,\frac{{2n + 7}}{{n + 2}}} \right| = \mathop {\lim }\limits_{n \to \infty } \frac{{\left( {n + 3} \right)\left( {2n + 7} \right)}}{{\left( {2n + 9} \right)\left( {n + 2} \right)}} = 1\]Shunga qaramay, nisbat testi bu erda bizga hech narsa aytmaydi. Biroq, biz tezda bu borada divergentsiya testidan foydalanishimiz mumkin. Aslida, bu bizning birinchi tanlovimiz bo'lishi kerak edi.
\[\mathop {\lim }\limits_{n \to \infty } \frac{{n + 2}}{{2n + 7}} = \frac{1}{2} \ne 0\]Divergensiya testiga ko'ra, bu qator divergent hisoblanadi.
Shunday qilib, oldingi ikkita misolda ko'rganimizdek, agar biz nisbat testidan \(L = 1\) ni olsak, seriya konvergent yoki divergent bo'lishi mumkin.
Keyingiga o'tishdan oldin nisbatlar testida e'tiborga olishimiz kerak bo'lgan yana bir narsa bor. Oxirgi qator polinomga bo'lingan ko'phad edi va biz nisbat testidan \(L = 1\) olganimizni ko'rdik. Bu har doim faqat polinomlar yoki radikallar ostidagi polinomlarni o'z ichiga olgan ratsional ifodalar bilan sodir bo'ladi. Shunday qilib, kelajakda bu kabi muammolar bo'yicha nisbat testini sinab ko'rishning hojati yo'q, chunki biz endi \(L = 1\) ni olishimizni bilamiz.
Bundan tashqari, ikkinchidan oxirgi misolda biz musbat atama polinomlarni o'z ichiga olgan ratsional ifoda bo'lgan o'zgaruvchan qator misolini ko'rdik va bu holatlarda biz har doim \(L = 1\) ni olamiz.
Keling, nisbat testining isboti bilan bo'limni yopamiz.
Nisbatan test isboti
Birinchidan, shuni yodda tutingki, biz umumiylikni yo'qotmasdan seriyalar \(n = 1\) da boshlanadi, deb taxmin qilishimiz mumkin, chunki biz barcha seriyalarimiz sinovlari uchun qilganmiz.
Keling, dalilni \(L < 1\) deb faraz qilishdan boshlaylik va biz \(\sum {{a_n}} \) mutlaqo konvergent ekanligini ko'rsatishimiz kerak. Buni amalga oshirish uchun avvalo shuni ta'kidlaymizki, \(L < 1\), chunki \(r\), masalan, \(L < r < 1\).
Endi eslang,
\[L = \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{a_{n + 1}}}}{{{a_n}}}} \right|\]va biz \(r\) ni tanlaganimiz sababli, \(L < r\) ba'zi \(N\) mavjud, agar \(n \ge N\) bizda bo'lsa,
\[\left| {\frac{{{a_{n + 1}}}}{{{a_n}}}} \right| < r\hspace{0.5in} \Rightarrow \hspace{0.5in}\left| {{a_{n + 1}}} \right| < r\left| {{a_n}} \right|\]Keyinchalik, quyidagilarni ko'rib chiqing,
\[\begin{align*}\left| {{a_{N + 1}}} \right| & < r\left| {{a_N}} \right|\\ \left| {{a_{N + 2}}} \right| & < r\left| {{a_{N + 1}}} \right| < {r^2}\left| {{a_N}} \right|\\ \left| {{a_{N + 3}}} \right| & < r\left| {{a_{N + 2}}} \right| < {r^3}\left| {{a_N}} \right|\\ & \hspace{0.5in} \vdots \\ \left| {{a_{N + k}}} \right| & < r\left| {{a_{N + k - 1}}} \right| < {r^k}\left| {{a_N}} \right|\end{align*}\]Shunday qilib, \(k = 1,2,3, \ldots \) uchun bizda \(\left| {{a_{N + k}}} \right| < {r^k}\left| {{a_N}} \right|\) mavjud. Faqat nima uchun bu muhim? Xo'sh, endi biz quyidagi seriyani ko'rishimiz mumkin.
\[\sum\limits_{k = 0}^\infty {\left| {{a_N}} \right|{r^k}} \]Bu geometrik qator va \(0 < r < 1\) tufayli biz aslida bu konvergent qator ekanligini bilamiz. Bundan tashqari, chunki \(\left| {{a_{N + k}}} \right| < {r^k}\left| {{a_N}} \right|\) taqqoslash orqali seriyani sinab ko'radi
\[\sum\limits_{n = N + 1}^\infty {\left| {{a_n}} \right|} = \sum\limits_{k = 1}^\infty {\left| {{a_{N + k}}} \right|} \]konvergent hisoblanadi. Biroq shundan beri,
\[\sum\limits_{n = 1}^\infty {\left| {{a_n}} \right|} = \sum\limits_{n = 1}^N {\left| {{a_n}} \right|} + \sum\limits_{n = N + 1}^\infty {\left| {{a_n}} \right|} \]biz \(\sum\limits_{n = 1}^\infty {\left| {{a_n}} \right|} \) ham yaqinlashishini bilamiz, chunki o'ngdagi birinchi had sonli hadlarning cheklangan yig'indisi va shuning uchun ham chekli. Shuning uchun \(\sum\limits_{n = 1}^\infty {{a_n}} \) mutlaqo konvergent hisoblanadi.
Keyinchalik, biz \(L > 1\) deb taxmin qilishimiz kerak va biz \(\sum {{a_n}} \) farqli ekanligini ko'rsatishimiz kerak. Shuni esga olib,
\[L = \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{a_{n + 1}}}}{{{a_n}}}} \right|\]va \(L > 1\) tufayli biz ba'zi \(N\) bo'lishi kerakligini bilamiz, agar \(n \ge N\) bizda bo'lsa,
\[\left| {\frac{{{a_{n + 1}}}}{{{a_n}}}} \right| > 1\hspace{0.5in} \Rightarrow \hspace{0.5in}\left| {{a_{n + 1}}} \right| > \left| {{a_n}} \right|\]Biroq, agar \(\left| {{a_{n + 1}}} \right| > \left| {{a_n}} \right|\) barcha \(n \ge N\) uchun bo'lsa, biz buni bilamiz,
\[\mathop {\lim }\limits_{n \to \infty } \left| {{a_n}} \right| \ne 0\]chunki shartlar kattalashib bormoqda va salbiy bo'lmasligi kafolatlangan. Bu o'z navbatida shuni anglatadiki,
\[\mathop {\lim }\limits_{n \to \infty } {a_n} \ne 0\]Shuning uchun, Divergentsiya testi bo'yicha \(\sum {{a_n}} \) divergent hisoblanadi.
Nihoyat, biz \(L = 1\) deb taxmin qilishimiz va uchta imkoniyatdan biriga ega bo'lgan seriyani olishimiz mumkinligini ko'rsatishimiz kerak. Buni amalga oshirish uchun bizga faqat har bir holat uchun seriya kerak. Tekshirish tafsilotlarini sizga qoldiramiz, ammo keyingi uchta seriyada \(L = 1\) mavjud va har birida imkoniyatlardan biri mavjud.
\[\begin{align*}& \sum\limits_{n = 1}^\infty {\frac{1}{{{n^2}}}} & \hspace{0.5in}&{\mbox{absolutely convergent}}\\ & \sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^n}}}{n}} & \hspace{0.5in}&{\mbox{conditionally convergent}}\\ & \sum\limits_{n = 1}^\infty {\frac{1}{n}} & \hspace{0.5in}&{\mbox{divergent}}\end{align*}\]