## Series

Let $X$ be a topological group with operation addition and let $\left({a}_{1},{a}_{2},{a}_{3},\dots \right)$ be a sequence in $X$.

• The series $\sum _{n=1}^{\infty }{a}_{n}$ is the sequence $\phantom{\rule{2em}{0ex}}\left({s}_{1},{s}_{2},{s}_{3},\dots \right)\phantom{\rule{2em}{0ex}}\text{where}\phantom{\rule{2em}{0ex}}{s}_{k}={a}_{1}+{a}_{2}+\dots +{a}_{k}.$
• Write $\sum _{n=1}^{\infty }{a}_{n}=a\phantom{\rule{2em}{0ex}}\text{if}\phantom{\rule{2em}{0ex}}\underset{n\to \infty }{\mathrm{lim}}{s}_{n}=a.$
• The series $\sum _{n=1}^{\infty }{a}_{n}$ converges if the sequence $\left({s}_{1},{s}_{2},{s}_{3},\dots \right)$ converges.
• The series $\sum _{n=1}^{\infty }{a}_{n}$ diverges if the sequence $\left({s}_{1},{s}_{2},{s}_{3},\dots \right)$ diverges.
• The series $\sum _{n=1}^{\infty }{a}_{n}$ converges absolutely if the series $\sum _{n=1}^{\infty }|{a}_{n}|$ converges.
• The series $\sum _{n=1}^{\infty }{a}_{n}$ converges conditionally if the series $\sum _{n=1}^{\infty }{a}_{n}$ converges and the series $\sum _{n=1}^{\infty }|{a}_{n}|$ diverges.

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

Let $\left({a}_{1},{a}_{2},{a}_{3},\dots \right)$ 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 $|r|<|s|$ then $\sum _{n=0}^{\infty }{a}_{n}{|r|}^{n}$ converges.
2. (b) If $\sum _{n=1}^{\infty }|{a}_{n}|$ converges then $\sum _{n=1}^{\infty }{a}_{n}$ converges.

 Proof of (a).

 Proof of (b). Let $\phantom{\rule{2em}{0ex}}{A}_{n}=|{a}_{1}|+|{a}_{2}|+\cdots +|{a}_{n}|\phantom{\rule{2em}{0ex}}\text{and}\phantom{\rule{2em}{0ex}}{s}_{n}={a}_{1}+{a}_{2}+\cdots +{a}_{n}.$ Since $\sum _{n=1}^{\infty }|{a}_{n}|=\left({A}_{1},{A}_{2},{A}_{3},\dots \right)$ converges, the sequence $\left({A}_{1},{A}_{2},{A}_{3},\dots \right)$ is Cauchy. Since $|{s}_{m}-{s}_{n}|=|{a}_{n+1}+\dots +{a}_{m}|\le |{a}_{m+1}|+\dots +|{a}_{n}|=|{A}_{m}-{A}_{n}|,$ the sequence $\left({s}_{1},{s}_{2},{s}_{3},\dots \right)$ is Cauchy. Since Cauchy sequences converge in $ℝ$ and $ℂ$ (in any complete metric space), the sequence $\left({s}_{1},{s}_{2},\dots \right)=\sum _{n=1}^{\infty }{a}_{n}$ converges. $\phantom{\rule{2em}{0ex}}\square$

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

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

 Proof. (a) Assume $\underset{n\to \infty }{\mathrm{lim}}\frac{{a}_{n+1}}{{a}_{n}}=a$ exists and $a<1$. Let $\epsilon \in {ℝ}_{>0}$ so that $a+\epsilon <1$. Since $\underset{n\to \infty }{\mathrm{lim}}\frac{{a}_{n+1}}{{a}_{n}}=a$, there exists $N\in {ℤ}_{>0}$ such that if $n\in {ℤ}_{>0}$ and $n>N$ then $\frac{{a}_{n+1}}{{a}_{n}}. Then $∑n=1∞ an = a1+a2 +⋯+aN+ aN+1+ aN+2+⋯ = a1+⋯+ aN+ aN+1+ aN+1 ( aN+2 aN+1 ) +aN+1 ( aN+2 aN+1 ) ( aN+3 aN+2 ) +⋯ < a1+⋯+ aN+ aN+1+ aN+1 (a+ε) +aN+1 (a+ε)2 +⋯ = a1+⋯+ aN+ aN+1 (1+ (a+ε) + (a+ε)2 +(a+ε)3 +⋯) = a1+⋯+ aN+ aN+1 ( 11-(a+ε) ).$ So $\sum _{n=1}^{\infty }{a}_{n}$ converges. (b) Assume $\underset{n\to \infty }{\mathrm{lim}}\frac{{a}_{n+1}}{{a}_{n}}=a$ exists and $a>1$. Let $\epsilon \in {ℝ}_{>0}$ so that $a-\epsilon <1$. Since $\underset{n\to \infty }{\mathrm{lim}}\frac{{a}_{n+1}}{{a}_{n}}$ there exists $N\in {ℤ}_{>0}$ such that if $n\in {ℤ}_{>0}$ and $n>N$ then $\frac{{a}_{n+1}}{{a}_{n}}>a-\epsilon$. Then $∑n=1∞ an = a1+a2 +⋯+aN+ aN+1+ aN+2+ ⋯ = a1+⋯+ aN+ aN+1+ aN+1 ( aN+2 aN+1) +aN+1 (aN+2 aN+1) (aN+3 aN+2) +⋯ > a1+⋯+ aN+ aN+1+ aN+1 (a-ε) +aN+1 (a-ε)2 +⋯ > a1+⋯+ aN+ aN+1 (1+1+1+⋯).$ So $\sum _{n+1}^{\infty }{a}_{n}$ diverges.

Leibniz's theorem. If $\left({a}_{1},{a}_{2},{a}_{3},\dots \right)$ is a sequence in $ℝ$ such that

1. ${a}_{n}\in {ℝ}_{\ge 0}$,
2. If $n\in {ℤ}_{>0}$ then ${a}_{n}\ge {a}_{n+1},$
3. If $\underset{n\to \infty }{\mathrm{lim}}{a}_{n}=0,$
then $\sum _{n=1}^{\infty }{\left(-1\right)}^{n-1}{a}_{n}$ converges.

 Proof.

1. (a) If $\left({a}_{1},{a}_{2},{a}_{3},\dots \right)$ is a sequence in $ℂ$, $\sum _{n=1}^{\infty }{a}_{n}=a$, and $\sum _{n=1}^{\infty }|{a}_{n}|$ converges absolutely
then every rearrangement of $\sum _{n=1}^{\infty }{a}_{n}$ converges to $a$.
2. (b) Assume that $\left({a}_{1},{a}_{2},{a}_{3},\dots \right)$ is a sequence in $ℝ$. Let $l\in ℝ$. If $\sum _{n=1}^{\infty }|{a}_{n}|$ 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 $k=1$, $∑n=1∞ 1n=1+ 12+( 13+14) +(15+ 16+17 +18)+⋯ > 1+12+ 12+12 +⋯$ So $\sum _{n=1}^{\infty }\frac{1}{n}$ diverges.

Example. $∑n=1∞ 1n2 = 1+( 122+ 132) +( 142+ 152+ 162+ 172) +(182 +…) < 1+222 +442+ 882 +… = 1+12+ 14+18 +… = 1+12+ (12)2 +(12)3 +… = 11-12 =2.$ In fact, according to Wolfram Alpha, $∑n=1∞ 1n2 =π26.$ If $k>1$ then $∑n=1∞ 1nk = 1+(12k +13k) +(14k+15k+ 16k+ 17k) +(18k +…) < 1+22k +44k+ 88k+⋯ = 1+1 2k-1 +14k-1 +18k-1 +⋯ = 1+ 12k-1 +( 12k-1 )2 +( 12k-1 )3 +⋯ = 11- 12k-1 = 2k-1 2k-1-1 .$ So $\sum _{n=1}^{\infty }\frac{1}{{n}^{k}}$ converges.

If $k<1$ then $∑n=1∞ 1nk = 1+12k +13k +14k +⋯ > 1+12 +13+14 +⋯.$ So $\sum _{n=1}^{\infty }\frac{1}{{n}^{k}}$ diverges.

Let $k\in {ℝ}_{>0}$. $∑n=1∞ 1nk converges if k>1, and ∑n=1∞ 1nk diverges if k≤1.$

Let $s\in ℂ$.

• The Riemann zeta function at $s$ is $ζ(s)= ∑n=1∞ 1ns.$

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?????.

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