For the series n=1 to infinity, 8/n^3, use the inequalities shown with n=10. Then, use the midpoint of the interval found to approximate the sum of the series.

Question

anonymous · Answer

$$s_{n}+\int\limits_{n+1}^{\infty} f(x)dx \le S \le s_{n}+ \int\limits_{n}^{\infty} f(x)dx$$

anonymous · Answer

a)$$? \le \sum_{n=1}^{\infty} \frac{ 8 }{ n^{3} }\le?$$

anonymous · Answer

@rational

rational · Answer

Yeah start by finding $s_n$

anonymous · Answer

It isn't a geometric or telescopic series right?

rational · Answer

use wolfram to find $s_{10}$

rational · Answer

$$s_{10} = \sum\limits_{n=1}^{10} \dfrac{8}{n^3} = ?$$

rational · Answer

http://www.wolframalpha.com/input/?i=%5Csum%5Climits_%7Bn%3D1%7D%5E%7B10%7D+%5Cdfrac%7B8%7D%7Bn%5E3%7D+

anonymous · Answer

9.580

rational · Answer

Yes so we have
$$\color{red}{9.580}+\int\limits_{10+1}^{\infty} f(x)dx \le S \le \color{red}{9.580}+ \int\limits_{10}^{\infty} f(x)dx$$

rational · Answer

plugin f(x) = 8/x^3 and evaluate the definite integrals above

anonymous · Answer

Alright, thanks!

rational · Answer

yw