10.5-bo'lim : Maxsus seriyalar
7. Qatlamning yaqinlashish yoki ayrilishligini aniqlang. Agar qator yaqinlashsa, uning qiymatini bering.
\[\sum\limits_{n = 4}^\infty {\frac{{10}}{{{n^2} - 4n + 3}}} \]Yechim
Ushbu bo'limda biz ko'rib chiqqan uchta maxsus seriyaning barchasi bir-biridan farqli ekanligini hisobga olsak, bu geometrik yoki garmonik qator emasligi aniq. Bu faqat teleskopni imkoniyat sifatida qoldiradi.
2-qadam
Endi bu erda ehtiyot bo'lishimiz kerak. Hozirgi vaqtda seriyani teleskopik seriya sifatida aniqlashning hech qanday usuli yo'q. Biz faqat teleskop seriyali deb umid qilamiz.
Shuning uchun, bu erda birinchi haqiqiy qadam, biz nimani olganimizni ko'rish uchun ketma-ket atama bo'yicha qisman kasrlarni bajarishdir. Bu yerda qator a'zosi uchun qisman kasr ish.
\[\frac{{10}}{{{n^2} - 4n + 3}} = \frac{{10}}{{\left( {n - 1} \right)\left( {n - 3} \right)}} = \frac{A}{{n - 1}} + \frac{B}{{n - 3}}\hspace{0.25in}\hspace{0.25in} \to \,\,\,\,\,\,\,\,\,\,\,10 = A\left( {n - 3} \right) + B\left( {n - 1} \right)\] \[\begin{array}{l}{n = 1}\hspace{0.5in}{10 = - 2A}\\{n = 3}\hspace{0.5in}{10 = 2B}\end{array}\hspace{0.25in} \to \hspace{0.25in}\hspace{0.25in}\begin{array}{l}{A = - 5}\\{B = 5}\end{array}\]Qisman kasr ko'rinishidagi qator atamasi u holda,
\[\frac{{10}}{{{n^2} - 4n + 3}} = \frac{5}{{n - 3}} - \frac{5}{{n - 1}}\]3-qadam
Ushbu seriya uchun qisman summalar,
\[{s_n} = \sum\limits_{i = 4}^n {\left[ {\frac{5}{{i - 3}} - \frac{5}{{i - 1}}} \right]} \]4-qadam
Oldingi bosqichdagi qisman summalarni kengaytirish,
\[\begin{align*}{s_n} & = \sum\limits_{i = 4}^n {\left[ {\frac{5}{{i - 3}} - \frac{5}{{i - 1}}} \right]} \\ & = \left[ {\frac{5}{1} - \require{cancel} \cancel{{\frac{5}{3}}}} \right] + \left[ {\frac{5}{2} - \require{cancel} \cancel{{\frac{5}{4}}}} \right] + \left[ {\require{cancel} \cancel{{\frac{5}{3}}} - \require{cancel} \bcancel{{\frac{5}{5}}}} \right] + \left[ {\require{cancel} \cancel{{\frac{5}{4}}} - \require{cancel} \bcancel{{\frac{5}{6}}}} \right] + \left[ {\require{cancel} \bcancel{{\frac{5}{5}}} - \require{cancel} \cancel{{\frac{5}{7}}}} \right] + \cdots \\ & \hspace{1.0in} + \left[ {\require{cancel} \cancel{{\frac{5}{{n - 7}}}} - \require{cancel} \bcancel{{\frac{5}{{n - 5}}}}} \right] + \left[ {\require{cancel} \bcancel{{\frac{5}{{n - 6}}}} - \require{cancel} \cancel{{\frac{5}{{n - 4}}}}} \right] + \left[ {\require{cancel} \bcancel{{\frac{5}{{n - 5}}}} - \require{cancel} \cancel{{\frac{5}{{n - 3}}}}} \right] + \\ & \hspace{1.5in}\left[ {\require{cancel} \cancel{{\frac{5}{{n - 4}}}} - \frac{5}{{n - 2}}} \right] + \left[ {\require{cancel} \cancel{{\frac{5}{{n - 3}}}} - \frac{5}{{n - 1}}} \right]\\ & = 5 + \frac{5}{2} - \frac{5}{{n - 2}} - \frac{5}{{n - 1}}\end{align*}\]Buni amalga oshirishda \(i\) ning boshlang‘ich va yakuniy qiymatlaridan kengayish va ketma-ket muddatning barcha qismlari bekor qilinmaguncha kengaytirish muhim ahamiyatga ega. Bu amalga oshirilgandan so'ng, bekor qilish kengayish tugaguniga qadar davom etadi deb taxmin qilish mumkin.
Bundan tashqari, yuqorida ko'rib turganingizdek, hamma narsa bekor qilinmaguncha kengayish juda chalkash bo'lishi mumkin, shuning uchun bu kabi tartibsizliklar paydo bo'lganda, bundan juda hayajonlanmang. Bu faqat ba'zida sodir bo'ladi va biz barcha kengayish bilan ehtiyot bo'lishimiz kerak.
E'tibor bering, hozir biz seriya teleskopik seriya ekanligini bilamiz, chunki biz bekor qilish uchun deyarli barcha "ichki" atamalarni oldik.
5-qadam
Shu nuqtada biz qilishimiz kerak bo'lgan narsa, olish uchun qisman summalarning chegarasiga qarashdir,
\[\mathop {\lim }\limits_{n \to \infty } {s_n} = \mathop {\lim }\limits_{n \to \infty } \left[ {\frac{{15}}{2} - \frac{5}{{n - 2}} - \frac{5}{{n - 1}}} \right] = \frac{{15}}{2}\]6-qadam
Qisman summalar chegarasi mavjud va cheklangan son (i.e. cheksizlik emas) va shuning uchun biz seriyani ko'rishimiz mumkin birlashadi va uning qiymati,
\[\sum\limits_{n = 4}^\infty {\frac{{10}}{{{n^2} - 4n + 3}}} = \frac{{15}}{2}\]