10.14-bo'lim : Quvvat seriyasi
1. Quyidagi daraja qatorlari uchun yaqinlashish oraliq va radiusini aniqlang.
\[\sum\limits_{n = 0}^\infty {\frac{1}{{{{\left( { - 3} \right)}^{2 + n}}\left( {{n^2} + 1} \right)}}{{\left( {4x - 12} \right)}^n}} \]Yechim
Mayli, keling, \(L\) bilan tanishish uchun nisbat testidan boshlaylik.
\[\begin{align*}L & = \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{{\left( {4x - 12} \right)}^{n + 1}}}}{{{{\left( { - 3} \right)}^{3 + n}}\left( {{{\left( {n + 1} \right)}^2} + 1} \right)}}\frac{{{{\left( { - 3} \right)}^{2 + n}}\left( {{n^2} + 1} \right)}}{{{{\left( {4x - 12} \right)}^n}}}} \right| = \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{\left( {4x - 12} \right)}}{{\left( { - 3} \right)\left( {{{\left( {n + 1} \right)}^2} + 1} \right)}}\frac{{\left( {{n^2} + 1} \right)}}{1}} \right|\\ & = \left| {4x - 12} \right|\mathop {\lim }\limits_{n \to \infty } \frac{{\left( {{n^2} + 1} \right)}}{{3\left( {{{\left( {n + 1} \right)}^2} + 1} \right)}} = \frac{1}{3}\left| {4x - 12} \right|\end{align*}\]2-qadam
Shunday qilib, biz bilamizki, seriyalar birlashadi, agar:
\[\frac{1}{3}\left| {4x - 12} \right| < 1\hspace{0.25in} \to \hspace{0.25in}\frac{4}{3}\left| {x - 3} \right| < 1\hspace{0.25in} \to \hspace{0.25in}\left| {x - 3} \right| < \frac{3}{4}\]3-qadam
Shunday qilib, oldingi bosqichdan biz yaqinlashish radiusi \(\require{bbox} \bbox[2pt,border:1px solid black]{{R = \frac{3}{4}}}\) ekanligini ko'ramiz.
4-qadam
Keling, konvergentsiya oralig'i ustida ishlashni boshlaylik. Keling, 2-bosqichda mavjud bo'lgan tengsizlikni yo'q qilaylik.
\[ - \frac{3}{4} < x - 3 < \frac{3}{4}\hspace{0.25in}\hspace{0.25in}\hspace{0.25in} \to \hspace{0.25in}\hspace{0.25in}\frac{9}{4} < x < \frac{{15}}{4}\]5-qadam
Konvergentsiya oralig'ini yakunlash uchun biz 4-bosqichdagi tengsizlikning oxirgi nuqtalarini tekshirishimiz kerak.
\(\displaystyle x = \frac{9}{4}\,\,\,:\,\,\,\sum\limits_{n = 0}^\infty {\frac{1}{{{{\left( { - 3} \right)}^{2 + n}}\left( {{n^2} + 1} \right)}}{{\left( { - 3} \right)}^n}} = \sum\limits_{n = 0}^\infty {\frac{1}{{{{\left( { - 3} \right)}^2}\left( {{n^2} + 1} \right)}}} = \sum\limits_{n = 0}^\infty {\frac{1}{{9\left( {{n^2} + 1} \right)}}} \)\(\displaystyle x = \frac{{15}}{4}\,\,\,:\,\,\,\sum\limits_{n = 0}^\infty {\frac{1}{{{{\left( { - 1} \right)}^{2 + n}}{{\left( 3 \right)}^{2 + n}}\left( {{n^2} + 1} \right)}}{{\left( 3 \right)}^n}} = \sum\limits_{n = 0}^\infty {\frac{1}{{{{\left( { - 1} \right)}^{2 + n}}{{\left( 3 \right)}^2}\left( {{n^2} + 1} \right)}}} = \sum\limits_{n = 0}^\infty {\frac{{{{\left( { - 1} \right)}^{2 + n}}}}{{9\left( {{n^2} + 1} \right)}}} \)
Endi biz birinchi seriyaning birlashishini ko'rish uchun tez taqqoslash testini o'tkazishimiz mumkin va ikkinchi seriyada tez o'zgaruvchan seriya testini o'tkazishimiz mumkin.
Bu ikkala bayonotni tekshirishni sizga qoldiramiz.
6-qadam
Konvergentsiya oralig'i quyida keltirilgan va xulosa qilish uchun yaqinlashuv radiusi ham ko'rsatilgan.
\[\require{bbox} \bbox[2pt,border:1px solid black]{{{\rm{Interval : }}\frac{9}{4} \le x \le \frac{{15}}{4}}}\hspace{0.25in}\hspace{0.25in}\hspace{0.25in}\require{bbox} \bbox[2pt,border:1px solid black]{{R = \frac{3}{4}}}\]