OK, biz ushbu quvvat seriyasi \(x = - 3\) uchun birlashishini bilamiz, ammo hozircha shunday. Biz konvergentsiyaga erishadigan \(x\)-ning qolgan qismini aniqlash uchun biz shu nuqtaga qadar muhokama qilgan har qanday testdan foydalanishimiz mumkin. Biz ishlashni tanlagan testni qo'llaganimizdan so'ng, biz \(x\) da shart(lar)ga erishamiz, ulardan quvvat seriyalari yaqinlashadigan \(x\) qiymatlarini va quvvat seriyalari ajralib chiqadigan \(x\) qiymatlarini aniqlash uchun foydalanishimiz mumkin. Bundan biz yaqinlashish radiusini va yaqinlashuv oralig'ining katta qismini olishimiz mumkin (oxirgi nuqtalardan tashqari).

Aytgancha, bu erda foydalanish uchun eng yaxshi testlar deyarli har doim nisbat yoki ildiz testidir. Biz ko'rib chiqadigan quvvat seriyalarining aksariyati bir yoki boshqasi uchun o'rnatilgan. Bunday holda biz nisbat testidan foydalanamiz.

\[\begin{align*}L & = \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{{\left( { - 1} \right)}^{n + 1}}\left( {n + 1} \right){{\left( {x + 3} \right)}^{n + 1}}}}{{{4^{n + 1}}}}\,\frac{{{4^n}}}{{{{\left( { - 1} \right)}^n}\left( n \right){{\left( {x + 3} \right)}^n}}}} \right|\\ & = \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{ - \left( {n + 1} \right)\left( {x + 3} \right)}}{{4n}}} \right|\end{align*}\]

Cheklov bilan uzoqroqqa borishdan oldin, \(x\) chegaraga bog'liq emasligi sababli, uni chegaradan chiqarib tashlash mumkinligiga e'tibor bering. Shuni ham yodda tutingki, buni amalga oshirishda biz mutlaq qiymat chiziqlarini saqlashimiz kerak, chunki biz hamma narsa ijobiy bo'lib qolishiga ishonch hosil qilishimiz kerak va \(x\) narsalarni salbiy qiladigan qiymat bo'lishi mumkin. Shunda chegara,

\[\begin{align*}L & = \left| {x + 3} \right|\mathop {\lim }\limits_{n \to \infty } \frac{{n + 1}}{{4n}}\\ & = \frac{1}{4}\left| {x + 3} \right|\end{align*}\]

Shunday qilib, nisbat testi shuni ko'rsatadiki, agar \(L < 1\) seriyalari yaqinlashsa, \(L > 1\) seriyalari ajralib chiqadi va \(L = 1\) bo'lsa, biz nima bo'lishini bilmaymiz. Demak, bizda bor,

\[\begin{align*}& \frac{1}{4}\left| {x + 3} \right| < 1\hspace{0.25in} \Rightarrow \hspace{0.25in}\left| {x + 3} \right| < 4\hspace{0.25in}\hspace{0.25in}{\mbox{series converges}}\\ & \frac{1}{4}\left| {x + 3} \right| > 1\hspace{0.25in} \Rightarrow \hspace{0.25in}\left| {x + 3} \right| > 4\hspace{0.25in}\hspace{0.25in}{\mbox{series diverges}}\end{align*}\]

Biz \(L = 1\) ishi bilan birozdan keyin shug'ullanamiz. E'tibor bering, bizda endi bu quvvat seriyasi uchun yaqinlashish radiusi mavjud. Aynan mana shu shartlar konvergentsiya radiusi uchun zarurdir. Ushbu quvvat seriyasi uchun yaqinlashish radiusi \(R = 4\).

Keling, konvergentsiya oralig'ini olaylik. Yuqoridagi birinchi tengsizlikni yechish orqali biz intervalning ko'p qismini (barchasi bo'lmasa) olamiz.

\[\begin{array}{c} - 4 < x + 3 < 4\\ - 7 < x < 1\end{array}\]

Shunday qilib, amal qilish oralig'ining katta qismi \( - 7 < x < 1\) tomonidan berilgan. Biz qilishimiz kerak bo'lgan yagona narsa, bu oraliqning so'nggi nuqtalarida quvvat seriyasining yaqinlashishi yoki ajralishini aniqlashdir. E'tibor bering, \(x\) ning ushbu qiymatlari \(L = 1\) ni beradigan \(x\) qiymatiga mos keladi.

Ushbu nuqtalarda yaqinlashuvni aniqlashning yo'li ularni oddiy quvvat seriyasiga ulash va kerakli test yordamida ketma-ketliklarning yaqinlashishi yoki ajralishini ko'rishdir.

\(x = - 7\):
Bu holda seriya,

\[\begin{align*}\sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^n}n}}{{{4^n}}}{{\left( { - 4} \right)}^n}} & = \sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^n}n}}{{{4^n}}}{{\left( { - 1} \right)}^n}{4^n}} \\ & = \sum\limits_{n = 1}^\infty {{{\left( { - 1} \right)}^n}{{\left( { - 1} \right)}^n}n} \hspace{0.25in}\hspace{0.25in}{\left( { - 1} \right)^n}{\left( { - 1} \right)^n} = {\left( { - 1} \right)^{2n}} = 1\\ & = \sum\limits_{n = 1}^\infty n \end{align*}\]

Bu seriya \(\mathop {\lim }\limits_{n \to \infty } n = \infty \ne 0\) dan beri Divergensiya testi bilan farqlanadi.

\(x = 1\):
Bu holda seriya,

\[\sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^n}n}}{{{4^n}}}{{\left( 4 \right)}^n}} = \sum\limits_{n = 1}^\infty {{{\left( { - 1} \right)}^n}n} \]

Bu qator Divergensiya testi bilan ham farqlanadi, chunki \(\mathop {\lim }\limits_{n \to \infty } {\left( { - 1} \right)^n}n\) mavjud emas.

Shunday qilib, bu holda quvvat seriyalari ikkala oxirgi nuqta uchun yaqinlashmaydi. U holda yaqinlashish oralig'i,

\[ - 7 < x < 1\]

Keling, to'g'ridan-to'g'ri nisbat testiga o'tamiz.

\[\begin{align*}L & = \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{2^{n + 1}}{{\left( {4x - 8} \right)}^{n + 1}}}}{{n + 1}}\,\,\frac{n}{{{2^n}{{\left( {4x - 8} \right)}^n}}}} \right|\\ & = \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{2n\left( {4x - 8} \right)}}{{n + 1}}} \right|\\ & = \left| {4x - 8} \right|\mathop {\lim }\limits_{n \to \infty } \frac{{2n}}{{n + 1}}\\ & = 2\left| {4x - 8} \right|\end{align*}\]

Shunday qilib, biz bundan quyidagi konvergentsiya/divergentsiya ma'lumotlarini olamiz.

\[\begin{align*}& 2\left| {4x - 8} \right| < 1\hspace{0.25in}\hspace{0.25in}{\mbox{series converges}}\\ & 2\left| {4x - 8} \right| > 1\hspace{0.25in}\hspace{0.25in}{\mbox{series diverges}}\end{align*}\]

Bu erda yaqinlashuv oralig'ini aniqlashda ehtiyot bo'lishimiz kerak. Konvergentsiya oralig'i \(\left| {x - a} \right| < R\) va \(\left| {x - a} \right| > R\) ni talab qiladi. Boshqacha qilib aytadigan bo'lsak, to'g'ri yaqinlashish radiusini olish uchun mutlaq qiymatlar qatoridan 4 ni koeffitsient qilishimiz kerak. Buni qilish,

\[\begin{align*} & 8\left| {x - 2} \right| < 1\hspace{0.25in} \Rightarrow \hspace{0.25in}\left| {x - 2} \right| < \frac{1}{8}\hspace{0.25in}\hspace{0.25in}{\mbox{series converges}}\\ & 8\left| {x - 2} \right| > 1\hspace{0.25in} \Rightarrow \hspace{0.25in}\left| {x - 2} \right| > \frac{1}{8}\hspace{0.25in}\hspace{0.25in}{\mbox{series diverges}}\end{align*}\]

Demak, bu darajalar qatori uchun yaqinlashish radiusi \(R = \frac{1}{8}\).

Keling, yaqinlashish oralig'ini topamiz. Shunga qaramay, biz birinchi navbatda yuqoridagi konvergentsiyani beradigan tengsizlikni hal qilamiz.

\[\begin{array}{c} - \frac{1}{8} < x - 2 < \frac{1}{8}\\ \frac{{15}}{8} < x < \frac{{17}}{8}\end{array}\]

Endi oxirgi nuqtalarni tekshiring.

\(\displaystyle x = \frac{{15}}{8}\):
Bu erda seriya,

\[\begin{align*}\sum\limits_{n = 1}^\infty {\frac{{{2^n}}}{n}{{\left( {\frac{{15}}{2} - 8} \right)}^n}} & = \sum\limits_{n = 1}^\infty {\frac{{{2^n}}}{n}{{\left( { - \frac{1}{2}} \right)}^n}} \\ & = \sum\limits_{n = 1}^\infty {\frac{{{2^n}}}{n}\frac{{{{\left( { - 1} \right)}^n}}}{{{2^n}}}} \\ & = \sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^n}}}{n}} \end{align*}\]

Bu o'zgaruvchan garmonik qator va biz bilamizki, u yaqinlashadi.

\(\displaystyle x = \frac{{17}}{8}\):
Bu erda seriya,

\[\begin{align*}\sum\limits_{n = 1}^\infty {\frac{{{2^n}}}{n}{{\left( {\frac{{17}}{2} - 8} \right)}^n}} & = \sum\limits_{n = 1}^\infty {\frac{{{2^n}}}{n}{{\left( {\frac{1}{2}} \right)}^n}} \\ & = \sum\limits_{n = 1}^\infty {\frac{{{2^n}}}{n}\frac{1}{{{2^n}}}} \\ & = \sum\limits_{n = 1}^\infty {\frac{1}{n}} \end{align*}\]

Bu garmonik seriya va biz uning ajralib turishini bilamiz.

Shunday qilib, quvvat seriyasi oxirgi nuqtalardan biri uchun birlashadi, ikkinchisi uchun emas. Bu tez-tez sodir bo'ladi, shuning uchun bu sodir bo'lganda hayajonlanmang. Bu darajalar qatori uchun yaqinlashish oralig'i,

\[\frac{{15}}{8} \le x < \frac{{17}}{8}\]

Biz ushbu misolni oldingi misollardagi kabi nisbat testidan boshlaymiz.

\[\begin{align*}L & = \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{\left( {n + 1} \right)!{{\left( {2x + 1} \right)}^{n + 1}}}}{{n!{{\left( {2x + 1} \right)}^n}}}} \right|\\ & = \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{\left( {n + 1} \right)n!\,\,\left( {2x + 1} \right)}}{{n!}}} \right|\\ & = \left| {2x + 1} \right|\mathop {\lim }\limits_{n \to \infty } \left( {n + 1} \right)\end{align*}\]

Bu vaqtda biz ehtiyot bo'lishimiz kerak. Cheklov cheksizdir, lekin chegara oldida \(x\) atamasi mavjud. Bizda \(L = \infty > 1\) mavjud \(x \ne - \frac{1}{2}\).

Shunday qilib, ushbu quvvat seriyasi faqat \(x = - \frac{1}{2}\) bilan birlashadi. Agar o'ylab ko'rsangiz, biz buni allaqachon bilganmiz. Dastlabki muhokamamizdan bilamizki, har bir quvvat seriyasi \(x = a\) va bu holda \(a = - \frac{1}{2}\) uchun birlashadi. Esda tutingki, biz \(a\)-ni \({\left( {x - a} \right)^n}\)-dan olamiz va \(x\) koeffitsienti bitta bo'lishi kerakligiga e'tibor bering!

Bu holda biz yaqinlashish radiusi \(R = 0\) va yaqinlashish oralig'i \(x = - \frac{1}{2}\) deb aytamiz va ha, biz haqiqatan ham yaqinlashuv oralig'ini nazarda tutdik, garchi bu faqat nuqta bo'lsa ham.

Ushbu misolda ildiz testi ko'proq mos keladi. Shunday qilib,

\[\begin{align*}L & = \mathop {\lim }\limits_{n \to \infty } {\left| {\frac{{{{\left( {x - 6} \right)}^n}}}{{{n^n}}}} \right|^{\frac{1}{n}}}\\ & = \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{x - 6}}{n}} \right|\\ & = \left| {x - 6} \right|\mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\\ & = 0\end{align*}\]

Shunday qilib, \(L = 0 < 1\) beri \(x\) qiymatidan qat'i nazar, bu quvvat seriyasi har bir \(x\) uchun birlashadi.

Bunday hollarda biz yaqinlashish radiusi \(R = \infty \) va yaqinlashish oralig'i \( - \infty < x < \infty \) deb aytamiz.

Avval ushbu muammoda \(a = 0\) ga e'tibor bering. Bu muammo uchun juda muhim emas, lekin odamlar bundan hayajonlanmasligi uchun buni ta'kidlash kerak.

Ushbu muammoning muhim farqi \(x\) dagi ko'rsatkichdir. Bu holda standart \(n\) emas, balki 2\(n\). Ko'rib turganimizdek, ba'zi quvvat seriyalarida \(n\) dan boshqa ko'rsatkichlar bo'ladi va shuning uchun biz hali ham bunday muammolarni hal qilishimiz kerak.

Bu yana ildiz testi uchun sozlanganga o'xshaydi, keling, undan foydalanaylik.

\[\begin{align*}L & = \mathop {\lim }\limits_{n \to \infty } {\left| {\frac{{{x^{2n}}}}{{{{\left( { - 3} \right)}^n}}}} \right|^{\frac{1}{n}}}\\ & = \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{x^2}}}{{ - 3}}} \right|\\ & = \frac{{{x^2}}}{3}\end{align*}\]

Shunday qilib, agar biz konvergentsiyaga erishamiz

\[\frac{{{x^2}}}{3} < 1\hspace{0.25in}\hspace{0.25in} \Rightarrow \hspace{0.25in}\hspace{0.25in}{x^2} < 3\]

Biroq, yaqinlashish radiusi 3 EMAS. Konvergentsiya radiusi \(x\) da 1 ko‘rsatkichini talab qiladi. Shuning uchun,

\[\begin{align*}\sqrt {{x^2}} & < \sqrt 3 \\ \left| x \right| & < \sqrt 3 \end{align*}\]

Mutlaq qiymat chiziqlari bilan ehtiyot bo'ling! Bu holda yaqinlashuv radiusi \(R = \sqrt 3 \) ga o'xshaydi. E'tibor bering, biz bu safar ajralish uchun tengsizlikni yo'q qilishdan bezovtalanmadik. Divergentsiya uchun tengsizlik bu shunchaki yaqinlashuv oralig'i bo'lib, test tengsizlik almashtirilganda beradi va umuman kerak emas. Biz odatda bu qismni o'tkazib yuboramiz.

Endi konvergentsiya oralig'ini olaylik. Birinchidan, biz tengsizlikdan olamiz,

\[ - \sqrt 3 < x < \sqrt 3 \]

Endi oxirgi nuqtalarni tekshiring.

\(x = - \sqrt 3 \):
Bu erda quvvat seriyasi,

\[\begin{align*}\sum\limits_{n = 1}^\infty {\frac{{{{\left( { - \sqrt 3 } \right)}^{2n}}}}{{{{\left( { - 3} \right)}^n}}}} & = \sum\limits_{n = 1}^\infty {\frac{{{{\left( {{{\left( { - \sqrt 3 } \right)}^2}} \right)}^n}}}{{{{\left( { - 3} \right)}^n}}}} \\ & = \sum\limits_{n = 1}^\infty {\frac{{{{\left( 3 \right)}^n}}}{{{{\left( { - 1} \right)}^n}{{\left( 3 \right)}^n}}}} \\ & = \sum\limits_{n = 1}^\infty {{{\left( { - 1} \right)}^n}} \end{align*}\]

\(\mathop {\lim }\limits_{n \to \infty } {\left( { - 1} \right)^n}\) mavjud emasligi sababli bu seriya Divergensiya testi bilan farqlanadi.

\(x = \sqrt 3 \):
Biz \(x\) ni kvadratga aylantirayotganimiz sababli, bu seriya avvalgi qadam bilan bir xil bo'ladi.

\[\sum\limits_{n = 1}^\infty {\frac{{{{\left( {\sqrt 3 } \right)}^{2n}}}}{{{{\left( { - 3} \right)}^n}}}} = \sum\limits_{n = 1}^\infty {{{\left( { - 1} \right)}^n}} \]

qaysi bir-biridan farq qiladi.

U holda yaqinlashish oralig'i,

\[ - \sqrt 3 < x < \sqrt 3 \]