Estimating the Value of a Series — Question 3

*Question 3 Let $\displaystyle S=\sum_{n=1}^{\infty}1/n^3$. Using the Integral Test upper remainder bound, find the least final index $N$ certified to give absolute error $|R_N|<\tfrac12 10^{-4}$. Explain whether this tolerance alone guarantees identical rounding to four decimal places.

*Question 3 – Solution

With $f(x)=x^{-3}$, $$0<R_N\le\int_N^{\infty}x^{-3}\,dx=\left[-\frac1{2x^2}\right]_N^{\infty}=\frac1{2N^2}.$$

The requested error is below $0.00005=5\times10^{-5}$: $$\frac1{2N^2}<5\times10^{-5}\iff N^2>10{,}000\iff N>100.$$ Thus the least integer certified by this bound is $\boxed{N=101}$.

At $N=100$, the bound equals $5\times10^{-5}$, so it does not guarantee a strict error smaller than half a unit in the fourth decimal place. At $N=101$, it does.

This absolute-error tolerance alone does not guarantee identical rounding: two nearby numbers can lie on opposite sides of a rounding threshold.