We seek an upper bound for the error of approximation of the series

${\displaystyle \underset{n=1}{\overset{}{}}}\frac{1}{n^3}$

by the partial sum of its $3$ first terms

${\displaystyle \underset{n=1}{\overset{3}{}}}\frac{1}{n^3}$

What is the right estimate?

The error is

${\displaystyle \underset{n=4}{\overset{}{}}}\frac{1}{n^3}{\displaystyle \underset{3}{\overset{}{}}}\frac{1}{x^3}dx=\frac{1}{2*3^2}.$

The error is

${\displaystyle \underset{n=3}{\overset{}{}}}\frac{1}{n^3}{\displaystyle \underset{2}{\overset{}{}}}\frac{1}{x^3}dx=\frac{1}{2*2^2}$

The error is

${\displaystyle \underset{n=3}{\overset{}{}}}\frac{1}{n^3}{\displaystyle \underset{3}{\overset{}{}}}\frac{1}{x^3}dx=\frac{1}{2*3^2}$

The error is

${\displaystyle \underset{n=4}{\overset{}{}}}\frac{1}{n^3}{\displaystyle \underset{4}{\overset{}{}}}\frac{1}{x^3}dx=\frac{1}{2*4^2}$