Chapter 1.16, Problem 30MP

1. It is clear that you (or your computer) can’t find the sum of an infinite series just by adding up the terms one by one. For example, to get $\varsigma (1.1)={\displaystyle {\sum }_{n=1}^{\infty }1/n^{1.1}}$
2. ( see Problem 15.22) with error < 0.005 takes about 10³³ terms. To see a simple alternative (for a series of positive decreasing terms) look at Figures 6.1 and 6.2. Show that when you have summed N terms, the sum R_N of the rest of the series is between $I_N={\displaystyle {\int }_N^{\infty }{a}_n}$ and $I_{N+1}={\displaystyle {\int }_{N+1}^{\infty }{a}_n}{d}_n$.

Find the integrals in (a) for the $\varsigma (1.1)$ series and verify the claimed number of terms needed for error Hint: Find N such that $I_N=0.005$. Also find upper and lower bounds for $\varsigma (1.1)$ by computing ${\displaystyle {\sum }_{n=1}^{N}1/n^{1.1}}+{\displaystyle {\int }_N^{\infty }{n}^{-1.1}}dn$ and ${\displaystyle {\sum }_{n=1}^{N}1/n^{1.1}}+{\displaystyle {\int }_{N+1}^{\infty }{n}^{-1.1}}dn$ where N is far less than 10³³. Hint: You want the difference the upper and lower limits to be about 0.005; find N so that term a_N = 0.005.