Sequences and Series

3.4 Sequences and Series

Sequences: $\left\{a_n\right\}\;n=0,1,2,\ldots\;\;a_n\in\mathbb{R}$ or $\mathbb{C}$ is called a sequence

\[\left\{a_n\right\}\stackrel{n\to\infty}{\longrightarrow} a \quad \mbox{or }\quad\lim_{n\to\infty} a_n=a\quad \mbox{if the sequence converges to a number $a\in\mathbb{R}$} \]

Series: $\left\{a_n\right\}$ sequence then $\sum_{n=0}^\infty a_n$ is called an infinite series

\[\begin{array}{cl}\sum_{n=0}^\infty a_n& \mbox{infinite series?}\\ \sum_{n=0}^N a_n& \mbox{finite series = o.k.}\\ \end{array}\]

If $S_N=\sum\limits_{n=0}^\infty a_n$ converges then this means that $\lim\limits_{N\to\infty} S_N = \lim\limits_{N\to\infty} \sum\limits_{n=0}^N a_n=\sum\limits_{n=0}^\infty a_n=\alpha$
condition: $a_n\rightarrow0$ if $\sum\limits_{n=1}^\infty \frac{1}{n}=\infty$!, hence $a_n\rightarrow0$ is not a sufficient condition for a series to converge.
Nearly all theoretical considerations related to the convergence conditions for series rely on the geometrical series:

\[\sum_{n=0}^\infty x^n;\qquad a_n=x^n\]

We can use a small trick to calculate the sum

\[S_N = \sum_{n=0}^N x^n.\]

Let us write the sum twice

\[\begin{array}{cccccclc} S_N & = & 1 + & x + & x^2 +& ... +& x^N & \\ x\;S_N & = & & x + & x^2 +& ... +& x^N +& x^{N+1} \end{array}\]

subtracting both lines we get

\[(1-x) \;S_N = 1 - x^{N+1} \quad \mbox{or} \quad S_N = \frac{1 - x^{N+1}}{1-x}\]

so for $\left|x\right|\lt1$ we find

\[\lim\limits_{N\to\infty} S_N = \frac{1}{1-x}\]

and for $\left|x\right|\rangle 1$ the series diverges.
For the geometric series we have two conditions for the components $a_n$

\[\frac{a_{n+1}}{a_n}=x\]
\[\sqrt[n]{a_n}=x\]

These properties allow to specify the conditions to take the geometric series as a majorant for other series leading to the two standard convergence criteria for infinite series:

If $\sum\limits_{n=0}^\infty a_n$ is given and $q$, $n_0$ exist with $\left|\frac{a_{n+1}}{a_n}\right|\leq q \lt1$ for $n\ge n_0$ then $\sum\limits_{n=0}^\infty a_n$ is convergent.
If $\left|\frac{a_{n+1}}{a_n}\right|\ge q\gt1$ then $\sum a_n$ is divergent, if $\left|\frac{a_{n+1}}{a_n}\right|\Rightarrow1$ no decision is possible.
If $q$, $n_0$ exist with $\sqrt[n]{a_n}\le q\lt1$ for $n\ge n_0$ then $\sum a_n$ convergent. If $\sqrt[n]{a_n}\ge q\ge1$ then $\sum a_n$ divergent. If $\sqrt[n]{a_n}\to 1$ no decision is possible.

3.4.1 Examples: Convergence of infinite series