Chapter 10.5, Problem 3E

To determine

To find: the six partial sums.

Explanation of Solution

Given:

  ${\displaystyle \sum _{n=1}^{\infty }\frac{1}{n}}$

Calculation:

Adding terms to the sum, starting from 1 up to 6

  $\begin{array}{c}{\displaystyle \sum _{n=1}^1\frac{1}{n}=\frac{1}{1}=1}\\ {\displaystyle \sum _{n=1}^2\frac{1}{n}=1+\frac{1}{2}=\frac{3}{2}}\\ {\displaystyle \sum _{n=1}^3\frac{1}{n}=\frac{3}{2}+\frac{1}{3}=\frac{11}{6}}\\ {\displaystyle \sum _{n=1}^4\frac{1}{n}=\frac{11}{6}+\frac{1}{4}=\frac{25}{12}}\\ {\displaystyle \sum _{n=1}^5\frac{1}{n}=\frac{25}{12}+\frac{1}{5}=\frac{137}{60}}\\ {\displaystyle \sum _{n=1}^6\frac{1}{n}=\frac{137}{60}+\frac{1}{6}=\frac{147}{60}=\frac{49}{20}}\end{array}$