10.13-bo'lim : Seriyaning qiymatini baholash
2. \(\displaystyle \sum\limits_{n = 3}^\infty {\frac{1}{{{n^3}\ln \left( n \right)}}} \) qiymatini baholash uchun Taqqoslash testi va \(n = 20\) dan foydalaning.
Yechim
Bizdan seriyaning qiymatini baholash uchun taqqoslash testidan foydalanishimiz so'ralganligi sababli, avvalo, taqqoslash testidan ushbu seriyada foydalanish mumkinligiga ishonch hosil qilishimiz kerak.
Bunday holda, bu juda oson, chunki bizning \(n\) diapazoni uchun seriyali atamalar aniq ijobiy va shuning uchun biz taqqoslash testidan foydalanishimiz mumkin.
Muammoni hal qilishdan oldin ushbu shartlarni sinab ko'rish juda muhimligini unutmang. Agar testni seriyada ishlatib bo'lmasa, seriyaning qiymatini baholash uchun testdan foydalanish mantiqiy emas. Bu erda bizdan testdan foydalanishni so'rashganligi sababli, test haqiqatdan ham ishlatilishi mumkin deb o'ylamasligimiz kerak!
2-qadam
Keling, \(n = 20\) yordamida qisman yig'indidan boshlaylik. Bu,
\[{s_{20}} = \sum\limits_{n = 3}^{20} {\frac{1}{{{n^3}\ln \left( n \right)}}} = 0.057315878\]3-qadam
Keling, seriya qiymatining ushbu yaqinlashuvida xato bahosini olishimiz mumkinligini bilib olaylik. Buni amalga oshirish uchun biz avval ushbu seriyada taqqoslash testini o'tkazishimiz kerak.
Agar biz \(\ln \left( n \right)\) ortib borayotgan funktsiya va \(\ln \left( n \right) \ge \ln \left( 3 \right)\) ekanligini payqasak, bu seriya uchun bu juda oson. Shunday qilib, biz olamiz,
\[\frac{1}{{{n^3}\ln \left( n \right)}} \le \frac{1}{{{n^3}\ln \left( 3 \right)}} = \frac{1}{{\ln \left( 3 \right)}}\frac{1}{{{n^3}}}\]4-qadam
Endi biz eslatmalardagi muhokamadan bilamizki, qolgan qiymatning yuqori chegarasi (i.e. taxminiy va aniq qiymat o'rtasidagi xatolik),
\[\begin{align*}{R_{20}} \le {T_{20}} &= \sum\limits_{n = 21}^\infty {\frac{1}{{{n^3}\ln \left( 3 \right)}}} < \int_{{20}}^{\infty }{{\frac{1}{{{x^3}\ln \left( 3 \right)}}\,dx}}\\ & = \mathop {\lim }\limits_{t \to \infty } \int_{{20}}^{t}{{\frac{1}{{{x^3}\ln \left( 3 \right)}}\,dx}} = \mathop {\lim }\limits_{t \to \infty } \left. {\left( { - \frac{1}{{2{x^2}\ln \left( 3 \right)}}} \right)} \right|_{20}^t\\ & = \mathop {\lim }\limits_{t \to \infty } \left( {\frac{1}{{800\ln \left( 3 \right)}} - \frac{1}{{2{t^2}\ln \left( 3 \right)}}} \right) = \frac{1}{{800\ln \left( 3 \right)}}\end{align*}\]5-qadam
Shunday qilib, biz seriyaning qiymatini taxmin qilishimiz mumkin,
\[\require{bbox} \bbox[2pt,border:1px solid black]{{s \approx 0.057315878}}\]va bu taxmindagi xato \(\frac{1}{{800\ln \left( 3 \right)}} = 0.001137799\) dan oshmaydi.