Series

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

Last updates: 31 May 2010

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) where s_{k} = a_{1} + a_{2} + \dots a_{k} .

Write

\sum_{n = 1}^{\infty} a_{n} = a if \lim_{n \to \infty} 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 $\sum_{n = 1}^{\infty} | a_{n} |$ 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.

Suppose that

\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 $a$ .
If $\sum_{n = 1}^{\infty} b_{n}$ is a series and $\sum_{n = 1}^{\infty} b_{n} = b$ then $\sum_{n = 1}^{\infty} a_{n} b_{n} = a b$ . IS THIS RIGHT? IT IS DIFFERENT FROM THE ORIGINAL NOTES.

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.

Proof.

Since $\sum_{n = 0}^{\infty} a_{n} s^{n}$ converges, $\lim_{n \to \infty} |a_{n} s^{n}| = 0 .$ Let $ϵ \in ℝ_{> 0}$ . Then there exists $N \in ℤ_{> 0}$ such that if $n \in ℤ_{> 0}$ and $n > N$ then $|a_{n} s^{n}| < ϵ .$ Then $\begin{matrix} \sum_{n = 0}^{\infty} |a_{n} r^{n}| & = & |a_{0}| + |a_{1} r| + \dots + |a_{N} r^{N}| + |a_{N + 1} r^{N + 1}| + \dots \\ = & |a_{0}| + \dots + |a_{N} r^{N}| + |a_{N + 1} s^{N + 1}| |\frac{r^{N + 1}}{s^{N + 1}}| + |a_{N + 2} s^{N + 2}| |\frac{r^{N + 2}}{s^{N + 2}}| + \dots \\ > & |a_{0}| + \dots + |a_{N} r^{N}| + ϵ |\frac{r^{N + 1}}{s^{N + 1}}| + ϵ |\frac{r^{N + 2}}{s^{N + 2}}| + \dots \\ = & |a_{0}| + \dots + |a_{N} r^{N}| + ε |\frac{r^{N + 1}}{s^{N + 1}}| (1 + |\frac{r}{s}| + |\frac{r^{2}}{s^{2}}| + \dots) \\ = & |a_{0}| + \dots + |a_{N} r^{N}| + ε |\frac{r^{N + 1}}{s^{N + 1}}| (\frac{1}{1 - \frac{|r|}{|s|}}) \end{matrix}$ So $\sum_{n = 0}^{\infty} |a_{n}| |r^{n}|$ converges. So $\sum_{n = 0}^{\infty} a_{n} r^{n}$ converges.

$□$

Assume that there is a sequence in $ℝ$ or $ℂ .$ If $\sum_{n = 1}^{\infty} |a_{n}|$ converges then $\sum_{n = 1}^{\infty} a_{n}$ converges.

		Proof.
	Let $A_{n} = \|a_{1}\| + \|a_{2}\| + \dots + \|a_{n}\|$ and $s_{n} = a_{1} + a_{2} + \dots + a_{n} .$ Since $\sum_{n = 1}^{\infty} \|a_{n}\| = (A_{1} A_{2} A_{3} \dots)$ converges the sequence is Cauchy. Since $\|s_{m} - s_{n}\| = \|a_{n + 1} + \dots + a_{m}\| \leq \|a_{m + 1}\| + \dots + \|a_{n}\| = \|A_{m} - A_{n}\|,$ the sequence $s_{n}$ is Cauchy. Since Cauchy sequences converge in $ℝ$ and $ℂ$ (or any complete metric space) $\sum_{n = 1}^{n} a_{n} = s_{n}$ converges. $□$

Let $(a_{n})$ be a sequence in $ℝ_{\geq 0} .$

Assume $\lim_{n \to \infty} (\frac{a_{n + 1}}{a_{n}}) = a$ exists and $a < 1 .$ Then $\sum_{n = 1}^{\infty} a_{n}$ converges.
Assume $\lim_{n \to \infty} (\frac{a_{n + 1}}{a_{n}}) = a$ exists and $a > 1 .$ Then $\sum_{n = 1}^{\infty} a_{n}$ diverges.

		Proof.
	Assume $\lim_{n \to \infty} \frac{a_{n + 1}}{a_{n}}$ exists and $a < 1 .$ Let $ε \in ℝ_{> 0}$ so that $a + ε < 1 .$ Since $\lim_{n \to \infty} \frac{a_{n + 1}}{a_{n}}$ there exists $N \in ℤ_{> 0}$ with $\frac{a_{n + 1}}{a_{n}} < a + ε$ if $n \in ℤ_{> 0}$ with $n > N .$ Then $\begin{array}{rcl} \sum_{n = 1}^{\infty} a_{n} & = & a_{1} + a_{2} + \dots + a_{N} + a_{N + 1} + a_{N + 2} \dots \\ = & a_{1} + \dots + a_{N} + a_{N + 1} + a_{N + 1} (\frac{a_{N + 2}}{a_{N + 1}}) + a_{N + 1} (\frac{a_{N + 2}}{a_{N + 1}}) (\frac{a_{N + 3}}{a_{N + 2}}) + \dots \\ < & a_{1} + \dots + a_{N} + a_{N + 1} + a_{N + 1} (a + ε) + a_{N + 1} {(a + ε)}^{2} + \dots \\ = & a_{1} + \dots + a_{N} + a_{N + 1} (1 + (a + ε) + {(a + ε)}^{2} + {(a + ε)}^{3} + \dots) \\ = & a_{1} + \dots + a_{N} + a_{N + 1} (\frac{1}{1 - (a + ε)}) . \end{array}$ So $\sum_{n = 1}^{\infty} a_{n}$ converges. Assume $\lim_{n \to \infty} \frac{a_{n + 1}}{a_{n}}$ exists and $a > 1 .$ Let $ε \in ℝ_{> 0}$ so that $a - ε < 1 .$ Since $\lim_{n \to \infty} \frac{a_{n + 1}}{a_{n}}$ there exists $N \in ℤ_{> 0}$ with $\frac{a_{n + 1}}{a_{n}} > a - ε$ if $n \in ℤ_{> 0}$ with $n > N .$ Then $\begin{array}{rcl} \sum_{n = 1}^{\infty} a_{n} & = & a_{1} + a_{2} + \dots + a_{N} + a_{N + 1} + a_{N + 2} \dots \\ = & a_{1} + \dots + a_{N} + a_{N + 1} + a_{N + 1} (\frac{a_{N + 2}}{a_{N + 1}}) + a_{N + 1} (\frac{a_{N + 2}}{a_{N + 1}}) (\frac{a_{N + 3}}{a_{N + 2}}) + \dots \\ > & a_{1} + \dots + a_{N} + a_{N + 1} + a_{N + 1} (a - ε) + a_{N + 1} {(a - ε)}^{2} + \dots \\ > & a_{1} + \dots + a_{N} + a_{N + 1} (1 + 1 + 1 + \dots) \\ = & a_{1} + \dots + a_{N} + a_{N + 1} (\frac{1}{1 - (a + ε)}) . \end{array}$ So $\sum_{n + 1}^{\infty} a_{n}$ diverges.

Leibniz's theorem If ${(a_{n})}_{n \in ℕ}$ is a sequence in $ℝ$ such that

$a_{n} \in ℝ_{\geq 0},$
if $n \in ℤ_{> 0}$ then $a_{n} \geq a_{n + 1},$
if $\lim_{n \to \infty} a_{n} = 0,$ then $\sum_{n = 1}^{\infty} {(- 1)}^{n - 1} a_{n}$ converges.

Proof.

Assume $(a_{n})$ is a sequence in $ℝ$ and $a_{n} \in ℝ_{\geq 0},$ and if $n \in ℤ_{> 0}$ then $a_{n} \geq a_{n + 1}$ and $\lim_{n \to \infty} a_{n} = 0 .$

To show: $\sum_{n = 0}^{\infty} {(- 1)}^{n - 1} a_{n} = a_{1} - a_{2} + a_{3} - a_{4} + a_{5} + \dots$ coverges.

Let $s_{2 m} = (a_{1} - a_{2}) + (a_{3} - a_{4}) + \dots + (a_{2 m - 1} - a_{2 m}) .$

Then $s_{2 m} \leq s_{2 (m + 1)}$

Since $s_{2 m} = a_{1} - (a_{2} - a_{3}) - (a_{4} - a_{5}) - \dots - (a_{2 m - 2} - a_{2 m - 1}) - a_{2 m},$ then $s_{2 m} \leq a_{1} .$

So the sequence $(s_{2} s_{4} s_{6} \dots)$ is increasing and bounded above. So $\lim_{n \to \infty} s_{2 m}$ exists.

Let $l = \lim_{m \to \infty} s_{2 m} .$

Since $s_{2 m + 1} = s_{2 m} + a_{2 m + 1}$ then $\begin{matrix} \lim_{m \to \infty} s_{2 m + 1} & = & \lim_{m \to \infty} s_{2 m} + \lim_{m \to \infty} a_{2 m + 1} \\ = & l + 0 = l . \end{matrix}$ So $\lim_{m \to \infty} s_{m} = l .$

$□$

Harmonic series and the Riemann zeta function

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

Example. $\begin{matrix} \sum_{n = 1}^{\infty} \frac{1}{n^{2}} & = & 1 + (\frac{1}{2^{2}} + \frac{1}{3^{2}}) + (\frac{1}{4^{2}} + \frac{1}{5^{2}} + \frac{1}{6^{2}} + \frac{1}{7^{2}}) + (\frac{1}{8^{2}} + \dots \\ < & 1 + \frac{2}{2^{2}} + \frac{4}{4^{2}} + \frac{8}{8^{2}} + \dots \\ = & 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \dots \\ = & 1 + \frac{1}{2} + {(\frac{1}{2})}^{2} + {(\frac{1}{2})}^{3} + \dots = \frac{1}{1 - \frac{1}{2}} = 2 . \end{matrix}$ In fact, according to Wolfram Alpha, $\sum_{n = 1}^{\infty} \frac{1}{n^{2}} = \frac{π^{2}}{6} .$

If $k > 1$ then $\begin{matrix} \sum_{n = 1}^{\infty} \frac{1}{n^{k}} & = & 1 + (\frac{1}{2^{k}} + \frac{1}{3^{k}}) + (\frac{1}{4^{k}} + \frac{1}{5^{k}} + \frac{1}{6^{k}} + \frac{1}{7^{k}}) + (\frac{1}{8^{k}} + \dots) \\ < & 1 + \frac{2}{2^{k}} + \frac{4}{4^{k}} + \frac{8}{8^{k}} + \dots \\ = & 1 + \frac{1}{2^{k - 1}} + \frac{1}{4^{k - 1}} + \frac{1}{8^{k - 1}} + \dots \\ = & 1 + \frac{1}{2^{k - 1}} + {(\frac{1}{2^{k - 1}})}^{2} + {(\frac{1}{2^{k - 1}})}^{3} + \dots \\ = & \frac{1}{1 - \frac{1}{2^{k - 1}}} = \frac{2^{k - 1}}{2^{k - 1} - 1} \end{matrix}$ so that $\sum_{n = 1}^{\infty} \frac{1}{n^{k - 1}}$ converges.

If $k < 1$ then $\begin{matrix} \sum_{n = 1}^{\infty} \frac{1}{n^{k}} & = & 1 + \frac{1}{2^{k}} + \frac{1}{3^{k}} + \frac{1}{4^{k}} + \dots \\ > & 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \dots \end{matrix}$ so that $\sum_{n = 1}^{\infty} \frac{1}{n^{k}}$ diverges.

Let $k \in ℝ_{> 0} .$ $\sum_{n = 1}^{\infty} \frac{1}{n^{k}} converges if k > 1,$ $\sum_{n = 1}^{\infty} \frac{1}{n^{k}} diverges if k \leq 1 .$

Let $s \in ℂ .$ The Riemann zeta function at $s$ is $ζ (s) = \sum_{n = 1}^{\infty} \frac{1}{n^{s}}$

Example. $ζ (2) = \frac{π^{2}}{6} .$

References [PLACEHOLDER]

[BG] A. Braverman and D. Gaitsgory, Crystals via the affine Grassmanian, Duke Math. J. 107 no. 3, (2001), 561-575; arXiv:math/9909077v2, MR1828302 (2002e:20083)

page history