10.14-bo'lim : Quvvat seriyasi
5. Quyidagi daraja qatorlari uchun yaqinlashish oralig'i va radiusini aniqlang.
\[\sum\limits_{n = 1}^\infty {\frac{{{6^n}}}{n}\,{{\left( {4x - 1} \right)}^{n - 1}}} \]Yechim
Mayli, keling, \(L\) bilan tanishish uchun nisbat testidan boshlaylik.
\[L = \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{6^{n + 1}}{{\left( {4x - 1} \right)}^n}}}{{n + 1}}\,\frac{n}{{{6^n}{{\left( {4x - 1} \right)}^{n - 1}}}}} \right| = \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{6n\left( {4x - 1} \right)}}{{n + 1}}} \right| = \left| {4x - 1} \right|\mathop {\lim }\limits_{n \to \infty } \frac{{6n}}{{n + 1}} = 6\left| {4x - 1} \right|\]2-qadam
Shunday qilib, biz bilamizki, seriyalar birlashadi, agar:
\[6\left| {4x - 1} \right| < 1\hspace{0.25in} \to \hspace{0.25in}24\left| {x - \frac{1}{4}} \right| < 1\hspace{0.25in} \to \hspace{0.25in}\left| {x - \frac{1}{4}} \right| < \frac{1}{{24}}\]3-qadam
Shunday qilib, oldingi bosqichdan biz yaqinlashish radiusi \(\require{bbox} \bbox[2pt,border:1px solid black]{{R = \frac{1}{{24}}}}\) ekanligini ko'ramiz.
4-qadam
Keling, konvergentsiya oralig'i ustida ishlashni boshlaylik. Keling, 2-bosqichda mavjud bo'lgan tengsizlikni yo'q qilaylik.
\[ - \frac{1}{{24}} < x - \frac{1}{4} < \frac{1}{{24}}\hspace{0.25in}\hspace{0.25in}\hspace{0.25in} \to \hspace{0.25in}\hspace{0.25in}\frac{5}{{24}} < x < \frac{7}{{24}}\]5-qadam
Konvergentsiya oralig'ini yakunlash uchun biz 4-bosqichdagi tengsizlikning oxirgi nuqtalarini tekshirishimiz kerak.
\(\displaystyle x = \frac{5}{{24}}\,\,\,:\,\,\,\sum\limits_{n = 1}^\infty {\frac{{{6^n}}}{n}\,{{\left( { - \frac{1}{6}} \right)}^{n - 1}}} = \sum\limits_{n = 1}^\infty {\frac{{{6^n}}}{n}\,\frac{{{{\left( { - 1} \right)}^{n - 1}}}}{{{6^{n - 1}}}}} = \sum\limits_{n = 1}^\infty {\frac{{6{{\left( { - 1} \right)}^{n - 1}}}}{n}} \)\(\displaystyle x = \frac{7}{{24}}\,\,\,:\,\,\,\sum\limits_{n = 1}^\infty {\frac{{{6^n}}}{n}\,{{\left( {\frac{1}{6}} \right)}^{n - 1}}} = \sum\limits_{n = 1}^\infty {\frac{{{6^n}}}{n}\,\frac{1}{{{6^{n - 1}}}}} = \sum\limits_{n = 1}^\infty {\frac{6}{n}} \)
Endi, birinchi seriya o'zgaruvchan garmonik seriya bo'lib, biz birlashadilar (yoki buni tekshirish uchun siz shunchaki muqobil seriyali testni o'tkazishingiz mumkin) va ikkinchi seriya \(p\) seriyali test orqali ajralib chiqadi.
6-qadam
Konvergentsiya oralig'i quyida keltirilgan va xulosa qilish uchun yaqinlashuv radiusi ham ko'rsatilgan.
\[\require{bbox} \bbox[2pt,border:1px solid black]{{{\rm{Interval : }}\frac{5}{{24}} \le x < \frac{7}{{24}}}}\hspace{0.25in}\hspace{0.25in}\hspace{0.25in}\require{bbox} \bbox[2pt,border:1px solid black]{{R = \frac{1}{{24}}}}\]