Series

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

Last updates: 2 June 2012

Series

Let $X$ be a topological group with operation addition and let $(a_1,a_2,a_3,\dots )$ be a sequence in $X$.

• The series $\sum _{n=1}^{\infty }{a}_n$ is the sequence $(s_1,s_2,s_3,\dots )\text{where}{s}_k=a_1+a_2+\dots +a_k.$
• Write $\sum _{n=1}^{\infty }{a}_n=a\text{if}\underset{n\to \infty }{\lim }{s}_n=a.$
• The series $\sum _{n=1}^{\infty }{a}_n$ converges if the sequence $(s_1,s_2,s_3,\dots )$ converges.
• The series $\sum _{n=1}^{\infty }{a}_n$ diverges if the sequence $(s_1,s_2,s_3,\dots )$ diverges.
• The series $\sum _{n=1}^{\infty }{a}_n$ converges absolutely if the series $\sum _{n=1}^{\infty }\left|a_n\right|$ converges.
• The series $\sum _{n=1}^{\infty }{a}_n$ converges conditionally if the series $\sum _{n=1}^{\infty }{a}_n$ converges and the series $\sum _{n=1}^{\infty }\left|a_n\right|$ diverges.

The following result establishes what is necessary to see that radius of convergence is a good notion for power series.

Let $(a_1,a_2,a_3,\dots )$ be a sequence in $ℝ$ or $ℂ$. Let $r,s\in ℂ$.

1. (a) Let $r,s\in ℂ$. Assume $\sum _{n=0}^{\infty }{a}_n{s}^n$ converges. If $\left|r\right|<\left|s\right|$ then $\sum _{n=0}^{\infty }{a}_n{\left|r\right|}^n$ converges.
2. (b) If $\sum _{n=1}^{\infty }\left|a_n\right|$ converges then $\sum _{n=1}^{\infty }{a}_n$ converges.

		Proof of (a).
	Since $\sum _{n=0}^{\infty }{a}_n{s}^n$ converges, $\underset{n\to \infty }{\lim }\left|a_n{s}^n\right|=0.$ Let $\epsilon \in {ℝ}_{>0}$. Then there exists $N\in {\mathbb{Z}}_{>0}$ such that if $n\in {\mathbb{Z}}_{>0}$ and $n>N$ then $\left|a_n{s}^n\right|<\epsilon .$ Then $$\begin{array}{rcl}\sum _{n=0}^{\infty }\left|a_n{r}^n\right|& =& \left|a_0\right|+\left|a_{1}r\right|+\cdots +\left|a_N{r}^N\right|+\left|a_{N+1}{r}^{N+1}\right|+\cdots \\ & =& \left|a_0\right|+\cdots +\left|a_N{r}^N\right|+\left|a_{N+1}{s}^{N+1}\right|\left|\frac{r^{N+1}}{s^{N+1}}\right|+\left|a_{N+2}{s}^{N+2}\right|\left|\frac{r^{N+2}}{s^{N+2}}\right|+\cdots \\ & <& \left|a_0\right|+\cdots +\left|a_N{r}^N\right|+\epsilon \left|\frac{r^{N+1}}{s^{N+1}}\right|+\epsilon \left|\frac{r^{N+2}}{s^{N+2}}\right|+\cdots \\ & =& \left|a_0\right|+\cdots +\left|a_N{r}^N\right|+\epsilon \left|\frac{r^{N+1}}{s^{N+1}}\right|(1+\left|\frac{r}{s}\right|+\left|\frac{r^2}{s^2}\right|+\cdots )\\ & =& \left|a_0\right|+\cdots +|a_N{r}^N|+\epsilon \left|\frac{r^{N+1}}{s^{N+1}}\right|\left(\frac{1}{1-\frac{\left|r\right|}{\left|s\right|}}\right)\end{array}$$ So $\sum _{n=0}^{\infty }\left|a_n\right|\left|r^n\right|$ converges. So $\sum _{n=0}^{\infty }{a}_n{r}^n$ converges. $\square$

		Proof of (b).
	Let $A_n=\left|a_1\right|+\left|a_2\right|+\cdots +\left|a_n\right|\text{and}{s}_n=a_1+a_2+\cdots +a_n.$ Since $\sum _{n=1}^{\infty }\left|a_n\right|=(A_1,A_2,A_3,\dots )$ converges, the sequence $(A_1,A_2,A_3,\dots )$ is Cauchy. Since $|s_m-s_n|=|a_{n+1}+\dots +a_m|\le \left|a_{m+1}\right|+\dots +\left|a_n\right|=|A_m-A_n|,$ the sequence $(s_1,s_2,s_3,\dots )$ is Cauchy. Since Cauchy sequences converge in $ℝ$ and $ℂ$ (in any complete metric space), the sequence $(s_1,s_2,\dots )=\sum _{n=1}^{\infty }{a}_n$ converges. $\square$

Let $\left(a_n\right)$ be a sequence in ${ℝ}_{\ge 0}$.

1. Assume $\underset{n\to \infty }{\lim }\left(\frac{a_{n+1}}{a_n}\right)=a$ exists and $a<1$. Then $\sum _{n=1}^{\infty }{a}_n$ converges.
2. Assume $\underset{n\to \infty }{\lim }\left(\frac{a_{n+1}}{a_n}\right)=a$ exists and $a>1$. Then $\sum _{n=1}^{\infty }{a}_n$ diverges.

		Proof.
	(a) Assume $\underset{n\to \infty }{\lim }\frac{a_{n+1}}{a_n}=a$ exists and $a<1$. Let $\epsilon \in {ℝ}_{>0}$ so that $a+\epsilon <1$. Since $\underset{n\to \infty }{\lim }\frac{a_{n+1}}{a_n}=a$, there exists $N\in {\mathbb{Z}}_{>0}$ such that if $n\in {\mathrm{<mo>\mathbb{Z}</mo>}}_{>0}$ and $n>N$ then $\frac{a_{n+1}}{a_n}<a+\epsilon$. Then $$\begin{array}{rcl}\sum _{n=1}^{\infty }{a}_n& =& a_1+a_2+\cdots +a_N+a_{N+1}+a_{N+2}+\cdots \\ & =& a_1+\cdots +a_N+a_{N+1}+a_{N+1}\left(\frac{a_{N+2}}{a_{N+1}}\right)+a_{N+1}\left(\frac{a_{N+2}}{a_{N+1}}\right)\left(\frac{a_{N+3}}{a_{N+2}}\right)+\cdots \\ & <& a_1+\cdots +a_N+a_{N+1}+a_{N+1}(a+\epsilon )+a_{N+1}{(a+\epsilon )}^2+\cdots \\ & =& a_1+\cdots +a_N+a_{N+1}(1+(a+\epsilon )+{(a+\epsilon )}^2+{(a+\epsilon )}^3+\cdots )\\ & =& a_1+\cdots +a_N+a_{N+1}\left(\frac{1}{1-(a+\epsilon )}\right).\end{array}$$ So $\sum _{n=1}^{\infty }{a}_n$ converges. (b) Assume $\underset{n\to \infty }{\lim }\frac{a_{n+1}}{a_n}=a$ exists and $a>1$. Let $\epsilon \in {ℝ}_{>0}$ so that $a-\epsilon <1$. Since $\underset{n\to \infty }{\lim }\frac{a_{n+1}}{a_n}$ there exists $N\in {\mathbb{Z}}_{>0}$ such that if $n\in {\mathrm{<mo>\mathbb{Z}</mo>}}_{>0}$ and $n>N$ then $\frac{a_{n+1}}{a_n}>a-\epsilon$. Then $$\begin{array}{rcl}\sum _{n=1}^{\infty }{a}_n& =& a_1+a_2+\cdots +a_N+a_{N+1}+a_{N+2}+\cdots \\ & =& a_1+\cdots +a_N+a_{N+1}+a_{N+1}\left(\frac{a_{N+2}}{a_{N+1}}\right)+a_{N+1}\left(\frac{a_{N+2}}{a_{N+1}}\right)\left(\frac{a_{N+3}}{a_{N+2}}\right)+\cdots \\ & >& a_1+\cdots +a_N+a_{N+1}+a_{N+1}(a-\epsilon )+a_{N+1}{(a-\epsilon )}^2+\cdots \\ & >& a_1+\cdots +a_N+a_{N+1}(1+1+1+\cdots ).\\ \end{array}$$ So $\sum _{n+1}^{\infty }{a}_n$ diverges.

Leibniz's theorem. If $(a_1,a_2,a_3,\dots )$ is a sequence in $ℝ$ such that

1. $a_n\in {ℝ}_{\ge 0}$,
2. If $n\in {\mathbb{Z}}_{>0}$ then $a_n\ge a_{n+1},$
3. If $\underset{n\to \infty }{\lim }{a}_n=0,$

then $\sum _{n=1}^{\infty }{(-1)}^{n-1}{a}_n$ converges.

		Proof.
	Assume $\left(a_n\right)$ is a sequence in $ℝ$ and $a_n\in {ℝ}_{\ge 0},$ and if $n\in {\mathbb{Z}}_{>0}$ then $a_n\ge a_{n+1}$ and $\underset{n\to \infty }{\lim }{a}_n=0$. To show: $\sum _{n=0}^{\infty }{(-1)}^{n-1}{a}_n=a_1-a_2+a_3-a_4+a_5-\cdots$ converges. Let $$s_{2m}=(a_1-a_2)+(a_3-a_4)+\cdots +(a_{2m-1}-a_{2m}).$$ Then $s_{2m}\le s_{2(m+1)}$ Since $s_{2m}=a_1-(a_2-a_3)-(a_4-a_5)-\cdots -(a_{2m-2}-a_{2m-1})-a_{2m}$, then $s_{2m}\le a_1$. So the sequence $(s_2,s_4,s_6,\dots )$ is increasing and bounded above. So $\underset{n\to \infty }{\lim }{s}_{2m}$ exists. Let $l=\underset{m\to \infty }{\lim }{s}_{2m}$. Since $s_{2m+1}=s_{2m}+a_{2m+1}$ then $$\underset{m\to \infty }{\lim }{s}_{2m+1}=\underset{m\to \infty }{\lim }{s}_{2m}+\underset{m\to \infty }{\lim }{a}_{2m+1}=l+0=l.$$ So $\underset{m\to \infty }{\lim }{s}_m=l$. $\square$

1. (a) If $(a_1,a_2,a_3,\dots )$ is a sequence in $ℂ$, $\sum _{n=1}^{\infty }{a}_n=a$, and $\sum _{n=1}^{\infty }\left|a_n\right|$ converges absolutely
   then every rearrangement of $\sum _{n=1}^{\infty }{a}_n$ converges to $a$.
2. (b) Assume that $(a_1,a_2,a_3,\dots )$ is a sequence in $ℝ$. Let $l\in ℝ$. If $\sum _{n=1}^{\infty }\left|a_n\right|$ converges conditionally (and not absolutely) then there exists a rearrangement of $\left(a_n\right)$ which converges to $l$.

Harmonic series and the Riemann zeta function

If $k=1$, $$\sum _{n=1}^{\infty }\frac{1}{n}=1+\frac{1}{2}+(\frac{1}{3}+\frac{1}{4})+(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8})+\cdots >1+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\cdots$$ So $\sum _{n=1}^{\infty }\frac{1}{n}$ diverges.

Example. $$\begin{array}{ccc}\sum _{n=1}^{\infty }\frac{1}{n^2}& =& 1+(\frac{1}{2^2}+\frac{1}{3^2})+(\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2})+(\frac{1}{8^2}+\dots )\\ & <& 1+\frac{2}{2^2}+\frac{4}{4^2}+\frac{8}{8^2}+\dots \\ & =& 1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\dots \\ & =& 1+\frac{1}{2}+{\left(\frac{1}{2}\right)}^2+\left(\frac{1}{2}\right)3+\dots =\frac{1}{1-\frac{1}{2}}=2.\end{array}$$ In fact, according to Wolfram Alpha, $$\sum _{n=1}^{\infty }\frac{1}{n^2}=\frac{{\pi }^2}{6}.$$ If $k>1$ then $$\begin{array}{ccc}\sum _{n=1}^{\infty }\frac{1}{n^k}& =& 1+(\frac{1}{2^k}+\frac{1}{3^k})+(\frac{1}{4^k}+\frac{1}{5^k}+\frac{1}{6^k}+\frac{1}{7^k})+(\frac{1}{8^k}+\dots )\\ & <& 1+\frac{2}{2^k}+\frac{4}{4^k}+\frac{8}{8^k}+\cdots \\ & =& 1+\frac{1}{2^{k-1}}+\frac{1}{4^{k-1}}+\frac{1}{8^{k-1}}+\cdots \\ & =& 1+\frac{1}{2^{k-1}}+{\left(\frac{1}{2^{k-1}}\right)}^2+{\left(\frac{1}{2^{k-1}}\right)}^3+\cdots \\ & =& \frac{1}{1-\frac{1}{2^{k-1}}}=\frac{2^{k-1}}{2^{k-1}-1}.\end{array}$$ So $\sum _{n=1}^{\infty }\frac{1}{n^k}$ converges.

If $k<1$ then $$\sum _{n=1}^{\infty }\frac{1}{n^k}=1+\frac{1}{2^k}+\frac{1}{3^k}+\frac{1}{4^k}+\cdots >1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots .$$ So $\sum _{n=1}^{\infty }\frac{1}{n^k}$ diverges.

Let $k\in {ℝ}_{>0}$. $$\sum _{n=1}^{\infty }\frac{1}{n^k}\text{converges if}k>1,\text{and}\sum _{n=1}^{\infty }\frac{1}{n^k}\text{diverges if}k\le 1.$$

Let $s\in ℂ$.

• The Riemann zeta function at $s$ is $$\zeta \left(s\right)=\sum _{n=1}^{\infty }\frac{1}{n^s}.$$

Example. $\zeta \left(2\right)=\frac{{\pi }^2}{6}$.

Notes and References

This section proves the popular series theorems for a undergraduate calculus course.

PERHAPS WE SHOULD STILL PUT IN INTEGRAL TESTS??? or are these too obvious, and should really just be done on examples????

A traditional reference for this material is [Ru1, Ch. 3 and 4] with Theorem (rearr) being [Ru1, Theorem 3.54 and 3.55], the Ratio test is [Ru1, Theorem 3.34], the Leibniz rule is (somewhat hidden) in [Ru1, Theorem 3.42 and Theorem 3.43]. The discussion of harmonic series is in [Ru1, Theorem 3.28].

Another traditional reference for this material is [TF, Chapter 16, particularly 16-5]. An advantage of this reference is that it has a wealth of examples.

PUT IN REFERENCES TO BOURBAKI.

WHERE WILL WE PUT SERIES EXAMPLES???

PUT IN A PROPER DEFINITION OF RADIUS OF CONVERGENCE.

References

[Bou] N. Bourbaki, Algèbre, Chapitre ?: ??????????? MR?????.

[Ru1] G.B. Thomas and R.L. Finney, Calculus and Analytic Geometry, Fifth edition, Addison Wesley, 1980. MR?????.

[Ru1] W. Rudin, Principles of mathematical analysis, McGraw-Hill, 1976. MR0385023

[Ru2] W. Rudin, Real and complex analysis, Third edition, McGraw-Hill, 1987. MR0924157.

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