10.6-bo'lim : Integral test
3. Quyidagi qatorlar yaqinlashadimi yoki ajraladimi, aniqlang.
\[\sum\limits_{n = 2}^\infty {\frac{1}{{{{\left( {2n + 7} \right)}^3}}}} \]Yechim
Xo'sh, ushbu seriyada Integral testdan foydalanishdan oldin biz Integral testdan foydalanishimiz mumkinligini tekshirishimiz kerak!
2-qadam
Seriya shartlari,
\[{a_n} = \frac{1}{{{{\left( {2n + 7} \right)}^3}}}\]Seriyadagi \(n\) diapazoni uchun shartlar ijobiy va shuning uchun shart bajarilganligini aniq ko'rishimiz mumkin.
3-qadam
Bunday holda, maxrajda faqat bitta \(n\) borligi va maxrajdagi barcha shartlar ijobiy bo'lgani uchun (umid qilamanki) aniq,
\[{a_n} = \frac{1}{{{{\left( {2n + 7} \right)}^3}}} > \frac{1}{{{{\left( {2\left( {n + 1} \right) + 7} \right)}^3}}} = {a_{n + 1}}\]va shuning uchun qator terminlari kamayib bormoqda.
OK, endi bilamizki, Integral testdan foydalanishimiz uchun talab qilinadigan ikkala shart ham tasdiqlangan, biz Integral testni davom ettirishimiz mumkin.
Sinovni amalda qo'llashdan oldin har doim ma'lum bir seriyali test shartlarini tekshirish juda muhimdir. Ko'pgina talabalar ketma-ket test bilan yo'l qo'yadigan eng katta xatolardan biri bu test shartlariga javob bermaydigan seriyadagi testdan foydalanish va shu sababli noto'g'ri javob olishdir!
4-qadam
Endi test uchun integralni hisoblaylik.
\[\begin{align*}\int_{2}^{\infty }{{\frac{1}{{{{\left( {2x + 7} \right)}^3}}}\,dx}} & = \mathop {\lim }\limits_{t \to \infty } \int_{2}^{t}{{\frac{1}{{{{\left( {2x + 7} \right)}^3}}}\,dx}} = \mathop {\lim }\limits_{t \to \infty } \left. {\left( { - \frac{1}{4}\frac{1}{{{{\left( {2x + 7} \right)}^2}}}} \right)} \right|_2^t\\ & = \mathop {\lim }\limits_{t \to \infty } \left( { - \frac{1}{4}\frac{1}{{{{\left( {2t + 7} \right)}^2}}} + \frac{1}{4}\frac{1}{{{{\left( {11} \right)}^2}}}} \right) = \frac{1}{{484}}\end{align*}\]5-qadam
Xo'sh, oxirgi bosqichdagi integral konvergent integraldir va shuning uchun Integral testda qator ham bo'lishi kerak. konvergent seriya.