10.13-bo'lim : Seriyaning qiymatini baholash
1. \(\displaystyle \sum\limits_{n = 1}^\infty {\frac{n}{{{{\left( {{n^2} + 1} \right)}^2}}}} \) qiymatini baholash uchun Integral Test va \(n = 10\) dan foydalaning.
Yechim
Bizdan seriyaning qiymatini baholash uchun Integral testdan foydalanishni so'rashganligi sababli, avvalo, Integral testni ushbu seriyada ishlatish mumkinligiga ishonch hosil qilishimiz kerak.
Birinchidan, ketma-ket shartlar aniq ijobiy, shuning uchun shart bajariladi.
Keling, quyidagi funktsiya bo'yicha kichik hisob-kitoblarni bajaramiz.
\[f\left( x \right) = \frac{x}{{{{\left( {{x^2} + 1} \right)}^2}}}\hspace{0.25in}\hspace{0.25in}f'\left( x \right) = \frac{{1 - 3{x^2}}}{{{{\left( {{x^2} + 1} \right)}^3}}}\]Funktsiyaning hosilasi \(x > \frac{1}{{\sqrt 3 }} = 0.5774\) uchun manfiy bo'ladi va shuning uchun funktsiya bu diapazonda kamayadi. Funktsiya va ketma-ket atamalar bir xil bo'lgani uchun biz \(n in our series.
Therefore, the conditions required to use the Integral Test are met! Note that it is really important to test these conditions before proceeding with the problem. It doesn’t make any sense to use a test to estimate the value of a series if the test can’t be used on the series. We shouldn’t just assume that because we are being asked to use a test here that the test can actually be used!
2-qadam
Let’s start off with the partial sum using \(n = 10\) diapazoni uchun ketma-ket atamalar kamayib borayotganini ham ko'rishimiz mumkin. Bu,
\[{s_{10}} = \sum\limits_{n = 1}^{10} {\frac{n}{{{{\left( {{n^2} + 1} \right)}^2}}}} = 0.392632317\]3-qadam
Endi oldingi bosqichdagi qisman yig'indining aniqligini oshirish uchun biz quyidagi ikkita integralning har biridan foydalanishimiz mumkinligini bilamiz.
\[\begin{align*}\int_{{10}}^{\infty }{{\frac{x}{{{{\left( {{x^2} + 1} \right)}^2}}}\,dx}} & = \mathop {\lim }\limits_{t \to \infty } \int_{{10}}^{t}{{\frac{x}{{{{\left( {{x^2} + 1} \right)}^2}}}\,dx}} = \mathop {\lim }\limits_{t \to \infty } \left. {\left[ { - \frac{1}{{2\left( {{x^2} + 1} \right)}}} \right]} \right|_{10}^t = \mathop {\lim }\limits_{t \to \infty } \left[ {\frac{1}{{202}} - \frac{1}{{2\left( {{t^2} + 1} \right)}}} \right] = \frac{1}{{202}}\\ \int_{{11}}^{\infty }{{\frac{x}{{{{\left( {{x^2} + 1} \right)}^2}}}\,dx}} & = \mathop {\lim }\limits_{t \to \infty } \int_{{11}}^{t}{{\frac{x}{{{{\left( {{x^2} + 1} \right)}^2}}}\,dx}} = \mathop {\lim }\limits_{t \to \infty } \left. {\left[ { - \frac{1}{{2\left( {{x^2} + 1} \right)}}} \right]} \right|_{11}^t = \mathop {\lim }\limits_{t \to \infty } \left[ {\frac{1}{{244}} - \frac{1}{{2\left( {{t^2} + 1} \right)}}} \right] = \frac{1}{{244}}\end{align*}\]4-qadam
Xo'sh, biz ushbu bo'limdagi eslatmalardan bilamizki, agar \(s\) seriyaning haqiqiy qiymatini bildirsa, u quyidagi diapazonda bo'lishi kerak.
\[\begin{align*}0.392632317 + \frac{1}{{244}} < \, & s < 0.392632317 + \frac{1}{{202}}\\ 0.396730678 < \, & s < 0.397582813 \end{align*}\]5-qadam
Va nihoyat, agar yuqoridagi ikkita raqamni o'rtacha hisoblasak, biz yaxshiroq baho olamiz,
\[\require{bbox} \bbox[2pt,border:1px solid black]{{s \approx 0.397156745}}\]