Series

Author: John J Weber III, PhD Corresponding Textbook Sections:

• Section 11.2 – Series

Expected Educational Results

• Objective 19–01: I can explain the difference between a series and a sequence.

• Objective 19–02: I can classify a series as geometric, $p$$p$-series, harmonic, or telescoping.

• Objective 19–03: I can explain what it means for a series to converge and diverge.

• Objective 19–04: I can explain why ${\displaystyle \underset{n\to \mathrm{\infty }}{lim}{a}_n=0}$$\displaystyle\lim_{n\rightarrow\infty}{a_n}=0$ is a necessary but not sufficient condition for series convergence.

• Objective 19–05: I can use the appropriate Series Convergence Test to determine if a series converges or diverges.

• Objective 19–06: I can compute the interval on which a power series converges.

• Objective 19–07: I can compute the sum of a convergent geometric series.

• Objective 19–08: I can compute the sum of a telescoping series.

• Objective 19–09: I can use a geometric series to rewrite a repeating decimal as a rational number.

Bloom’s Taxonomy

A modern version of Bloom’s Taxonomy is included here to recognize various different levels of understanding and to encourage you to work towards higher-order understanding (those at the top of the pyramid). All Objectives, Investigations, Activities, etc. are color-coded with the level of understanding.

Series

A Proof of the Divergence of the Harmonic Series

NOTE: You will not be assessed on the proof of the divergence of the harmonic series.

Pietro Mengoli's (circa ${17}^{\text{th}}$$17^{\text{th}}$ century) proof of divergence of the harmonic series. We will use a Proof by Contradiction.

Suppose the harmonic series converges to $S$$S$, i.e.,

${\displaystyle \sum _{n=1}^{\mathrm{\infty }}\frac{1}{n}=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\cdots =S}$$\displaystyle\sum_{n=1}^{\infty}{\frac{1}{n}}=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\cdots=S$.

We will show this results in a contradiction and we will be able to conclude that the supposition statement is FALSE, i.e., the harmonic series diverges.

So,

${\displaystyle S=1+(\frac{1}{2}+\frac{1}{3}+\frac{1}{4})+(\frac{1}{5}+\frac{1}{6}+\frac{1}{7})+(\frac{1}{8}+\frac{1}{9}+\frac{1}{10})+(\frac{1}{11}+\frac{1}{12}+\frac{1}{13})+\dots }$$\displaystyle S=1+\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right)+\left(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}\right)+\left(\frac{1}{8}+\frac{1}{9}+\frac{1}{10}\right)+\left(\frac{1}{11}+\frac{1}{12}+\frac{1}{13}\right)+\dots$

but,

$\begin{array}{c}{\displaystyle \frac{1}{2}}+{\displaystyle \frac{1}{4}}>{\displaystyle \frac{2}{3}}\\ {\displaystyle \frac{1}{5}}+{\displaystyle \frac{1}{7}}>{\displaystyle \frac{2}{6}}\\ {\displaystyle \frac{1}{8}}+{\displaystyle \frac{1}{10}}>{\displaystyle \frac{2}{9}}\\ {\displaystyle \frac{1}{11}}+{\displaystyle \frac{1}{13}}>{\displaystyle \frac{2}{12}}\\ {\displaystyle \frac{1}{n-1}}+{\displaystyle \frac{1}{n+1}}>{\displaystyle \frac{2}{n}}\end{array}$$\dfrac{1}{2}+\dfrac{1}{4} > \dfrac{2}{3} \\ \dfrac{1}{5}+\dfrac{1}{7} > \dfrac{2}{6} \\ \dfrac{1}{8}+\dfrac{1}{10} > \dfrac{2}{9} \\ \dfrac{1}{11}+\dfrac{1}{13} > \dfrac{2}{12} \\ \dfrac{1}{n-1}+\dfrac{1}{n+1} > \dfrac{2}{n}$

and

$\begin{array}{c}{\displaystyle \frac{1}{2}}+{\displaystyle \frac{1}{3}}+{\displaystyle \frac{1}{4}}>{\displaystyle \frac{2}{3}}+{\displaystyle \frac{1}{3}}={\displaystyle \frac{3}{3}}\\ {\displaystyle \frac{1}{5}}+{\displaystyle \frac{1}{6}}+{\displaystyle \frac{1}{7}}>{\displaystyle \frac{2}{6}}+{\displaystyle \frac{1}{6}}={\displaystyle \frac{3}{6}}\\ {\displaystyle \frac{1}{8}}+{\displaystyle \frac{1}{9}}+{\displaystyle \frac{1}{10}}>{\displaystyle \frac{2}{9}}+{\displaystyle \frac{1}{9}}={\displaystyle \frac{3}{9}}\\ {\displaystyle \frac{1}{11}}+{\displaystyle \frac{1}{12}}+{\displaystyle \frac{1}{13}}>{\displaystyle \frac{2}{12}}+{\displaystyle \frac{1}{12}}={\displaystyle \frac{3}{12}}\\ {\displaystyle \frac{1}{n-1}}+{\displaystyle \frac{1}{n}}+{\displaystyle \frac{1}{n+1}}>{\displaystyle \frac{2}{n}}+{\displaystyle \frac{1}{n}}={\displaystyle \frac{3}{n}}\end{array}$$\textcolor{red}{\dfrac{1}{2}}+\dfrac{1}{3}+\textcolor{red}{\dfrac{1}{4}} > \textcolor{red}{\dfrac{2}{3}}+\dfrac{1}{3} = \dfrac{3}{3}\\ \textcolor{red}{\dfrac{1}{5}}+\dfrac{1}{6}+\textcolor{red}{\dfrac{1}{7}} > \textcolor{red}{\dfrac{2}{6}}+\dfrac{1}{6} = \dfrac{3}{6}\\ \textcolor{red}{\dfrac{1}{8}}+\dfrac{1}{9}+\textcolor{red}{\dfrac{1}{10}} > \textcolor{red}{\dfrac{2}{9}}+\dfrac{1}{9} = \dfrac{3}{9}\\ \textcolor{red}{\dfrac{1}{11}}+\dfrac{1}{12}+\textcolor{red}{\dfrac{1}{13}} > \textcolor{red}{\dfrac{2}{12}}+\dfrac{1}{12} = \dfrac{3}{12}\\ \textcolor{red}{\dfrac{1}{n-1}}+\dfrac{1}{n}+\textcolor{red}{\dfrac{1}{n+1}} > \textcolor{red}{\dfrac{2}{n}}+\dfrac{1}{n} = \dfrac{3}{n}$

So,

$\begin{array}{c}S=1+\left({\displaystyle \frac{1}{2}}+{\displaystyle \frac{1}{3}}+{\displaystyle \frac{1}{4}}\right)+\left({\displaystyle \frac{1}{5}}+{\displaystyle \frac{1}{6}}+{\displaystyle \frac{1}{7}}\right)+\left({\displaystyle \frac{1}{8}}+{\displaystyle \frac{1}{9}}+{\displaystyle \frac{1}{10}}\right)+\left({\displaystyle \frac{1}{11}}+{\displaystyle \frac{1}{12}}+{\displaystyle \frac{1}{13}}\right)+\dots \\ >1+{\displaystyle \frac{3}{3}}+{\displaystyle \frac{3}{6}}+{\displaystyle \frac{3}{9}}+{\displaystyle \frac{3}{12}}+\dots \\ =1+{\displaystyle \frac{3}{3}}\left(1+{\displaystyle \frac{1}{2}}+{\displaystyle \frac{1}{3}}+{\displaystyle \frac{1}{4}}+\dots \right)\\ =1+1S\text{ from the supposition statement}.\end{array}$$S = 1+\left(\textcolor{blue}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}}\right)+\left(\textcolor{green}{\dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}}\right)+\left(\textcolor{orange}{\dfrac{1}{8}+\dfrac{1}{9}+\dfrac{1}{10}}\right)+\left(\textcolor{brown}{\dfrac{1}{11}+\dfrac{1}{12}+\dfrac{1}{13}}\right)+\dots \\ \gt 1+\textcolor{blue}{\dfrac{3}{3}}+\textcolor{green}{\dfrac{3}{6}}+\textcolor{orange}{\dfrac{3}{9}}+\textcolor{brown}{\dfrac{3}{12}}+\dots \\ = 1+\dfrac{3}{3}\left(\textcolor{purple}{1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dots}\right) \\ = 1+1\textcolor{purple}{S} \text{ from the supposition statement}.$

As a result of the above, $S>1+S$$S>1+S$, which is a contradiction. Therefore, harmonic series are divergent.

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Last Modified: Thursday, 12 November 2020 6:42 EDT