10.10-bo'lim : Nisbat sinovi
5. Quyidagi qatorlar yaqinlashadimi yoki ayiriladimi, aniqlang.
\[\sum\limits_{n = 1}^\infty {\frac{{{{\left( { - 1} \right)}^{n + 1}}}}{{6n + 7}}} \]Yechim
Biz \(L\) ni hisoblashimiz kerak.
\[\begin{align*}L & = \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{a_{n + 1}}}}{{{a_n}}}} \right| = \mathop {\lim }\limits_{n \to \infty } \left| {{a_{n + 1}}\frac{1}{{{a_n}}}} \right| = \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{{\left( { - 1} \right)}^{n + 1 + 1}}}}{{6\left( {n + 1} \right) + 7}}\frac{{6n + 7}}{{{{\left( { - 1} \right)}^{n + 1}}}}} \right|\\ & = \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{{\left( { - 1} \right)}^{n + 2}}}}{{6n + 13}}\frac{{6n + 7}}{{{{\left( { - 1} \right)}^{n + 1}}}}} \right| = \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{\left( { - 1} \right)\left( {6n + 7} \right)}}{{6n + 13}}} \right| = 1\end{align*}\]\({a_{n + 1}}\) ni hisoblashda \(n\) koeffitsientlariga va \(n\) quvvatlariga e'tibor bering. Bular bilan to'g'ri shug'ullanmaslik talabalarning ushbu hisoblashda eng katta xatolaridan biridir va bu darajadagi xatolar ko'pincha noto'g'ri javobga olib keladi!
2-qadam
OK, biz \(L = 1\) va shunga o'xshash nisbat testi bu seriya haqida hech narsa aytmasligini ko'rishimiz mumkin.
3-qadam
Nisbatan test bizga hech narsa aytmasa, bu seriyaning yaqinlashish yoki ajralishini aniqlay olmasligimizni anglatmaydi.
Aslida, bu holatda buni qilish juda oddiy. Bu muqobil seriyadir,
\[{b_n} = \frac{1}{{6n + 7}}\]\({b_n}\) aniq ijobiy va ular ham kamayib borayotgan ketma-ketlikni tashkil etishi juda aniq (umid qilamanki) bo'lishi kerak. Va nihoyat, biz \(\mathop {\lim }\limits_{n \to \infty } {b_n} = 0\) va shunga o'xshash muqobil seriyali test orqali ushbu seriyani ko'rishimiz mumkin. birlashish.
E'tibor bering, agar ushbu seriya ushbu bo'limda bo'lmaganida, boshidan muqobil seriya sifatida qilish bu muammoni hal qilishning eng yaxshi usuli bo'lar edi.