Chapter 9.5, Problem 3E

To determine

To find: First six partial sum of the series ${\displaystyle \sum _{n=1}^{\infty }\frac{1}{n}}$ .

Answer to Problem 3E

The first six partial sum of ${\displaystyle \sum _{n=1}^{\infty }\frac{1}{n}}$ are $S_1=1$ , $S_2=\frac{3}{2}$ , $S_3=\frac{11}{6}$ , $S_4=\frac{25}{12}$ , $S_5=\frac{137}{60}$ , and $S_6=\frac{49}{20}$ .

Explanation of Solution

Given information: The series ${\displaystyle \sum _{n=1}^{\infty }\frac{1}{n}}$ is given.

Concept used :

The partial sum of an infinite series is determined by the sum of a finite number of consecutive terms starting with the first term. 

The first partial sum is just the first term while the second partial sum is the sum of the first two terms. Similarly, the third partial sum is the sum of the first three terms and so on.

Calculation:

Substitute $n=1,2,3,4,5,6$ into the given series and calculate the corresponding partial sum.

First partial sum:

  $\begin{array}{c}{S}_1=\frac{1}{1}\\ =1\end{array}$

Second partial sum:

  $\begin{array}{c}{S}_2={\displaystyle \sum _{n=1}^2\frac{1}{n}}\\ =\frac{1}{1}+\frac{1}{2}\\ =\frac{2+1}{2}\\ =\frac{3}{2}\end{array}$

Third partial sum:

  $\begin{array}{c}{S}_3={\displaystyle \sum _{n=1}^3\frac{1}{n}}\\ =\frac{1}{1}+\frac{1}{2}+\frac{1}{3}\\ =\frac{6+3+2}{6}\\ =\frac{11}{6}\end{array}$

Fourth partial sum:

  $\begin{array}{c}{S}_4={\displaystyle \sum _{n=1}^4\frac{1}{n}}\\ =\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\\ =\frac{12+6+4+3}{12}\\ =\frac{25}{12}\end{array}$

Fifth partial sum:

  $\begin{array}{c}{S}_5={\displaystyle \sum _{n=1}^5\frac{1}{n}}\\ =\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}\\ =\frac{60+30+20+15+12}{60}\\ =\frac{137}{60}\end{array}$

Sixth partial sum:

  $\begin{array}{c}{S}_6={\displaystyle \sum _{n=1}^6\frac{1}{n}}\\ =\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}\\ =\frac{60+30+20+15+12+10}{60}\\ =\frac{147}{60}\\ =\frac{49}{20}\end{array}$

Therefore, the first six partial sum of the given series are as follows:

  $S_1=1$

  $S_2=\frac{3}{2}$

  $S_3=\frac{11}{6}$

  $S_4=\frac{25}{12}$ ,

  $S_5=\frac{137}{60}$

  $S_6=\frac{49}{20}$