MATH 152 Chapter 10.2

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Series

Definitions

Term

$a_{n}$

$N$^th Partial Sum

$\sum _{n=1}^{N}{a_{n}}=a_{1}+a_{2}+a_{3}+\dots +a_{N-1}+a_{N}$

Infinite Sum

$\sum _{n=1}^{\infty }{a_{n}}=a_{1}+a_{2}+a_{3}+\dots +a_{n-1}+a_{n}+\dots$

Geometric Series

Converges only if $\left|r\right|<1$

${\begin{array}{rcl}\sum _{n=1}^{\infty }{ar^{n-1}}=s_{n}&=&a+ar+ar^{2}+ar^{3}+\dots +ar^{n-2}+ar^{n-1}\\rs_{n}&=&ar+ar^{2}+ar^{3}+ar^{4}+\dots +ar^{n-1}+ar^{n}\\s_{n}-rs_{n}&=&a+ar^{n}\\s_{n}&=&{\frac {a\left(1+r^{n}\right)}{1-r}}\quad {\mbox{if}}\;\left|r\right|<1\;{\mbox{, then}}\;\lim _{n\to \infty }r^{n}\to 0\\\sum _{n=1}^{\infty }{ar^{n-1}}=\lim _{n\to \infty }{\frac {a\left(1+r^{n}\right)}{1-r}}&=&{\frac {a}{1-r}}\quad \left|r\right|<1\end{array}}$

Telescoping Series

${\begin{array}{rcl}\sum _{n=1}^{\infty }{\left({\frac {1}{n}}-{\frac {1}{n+3}}\right)}=s_{N}&=&\left(1-{\frac {1}{4}}\right)+\left({\frac {1}{2}}-{\frac {1}{5}}\right)+\left({\frac {1}{3}}-{\frac {1}{6}}\right)+\left({\frac {1}{4}}-{\frac {1}{7}}\right)+\left({\frac {1}{5}}-{\frac {1}{8}}\right)+\dots +\\&&\left({\frac {1}{N-2}}+{\frac {1}{N+1}}\right)+\left({\frac {1}{N-1}}-{\frac {1}{N+2}}\right)+\left({\frac {1}{N}}-{\frac {1}{N+3}}\right)\\&=&1+{\frac {1}{2}}+{\frac {1}{3}}-{\frac {1}{N+1}}-{\frac {1}{N+2}}-{\frac {1}{N+3}}\\\lim _{N\to \infty }s_{N}&=&{\frac {11}{6}}\end{array}}$

NOTE: $a_{n}={\tfrac {3}{n\left(n+3\right)}}$; use partial fractions to get back to standard form

Properties (if convergent)

If $a_{n}$ and $b_{n}$ are convergent, then...

${\begin{array}{rcl}\sum _{n=1}^{\infty }{\left(a_{n}\pm b_{n}\right)}&=&\sum _{n=1}^{\infty }{a_{n}}\pm \sum _{n=1}^{\infty }{b_{n}}\\\sum _{n=1}^{\infty }{ca_{n}}&=&c\sum _{n=1}^{\infty }{a_{n}}\end{array}}$

Tests for Convergence

If $\sum {a_{n}}$ is convergent, then $\lim _{n\to \infty }{a_{n}}=0$

Contrapositive: If $\lim _{n\to \infty }{a_{n}}\neq 0$, then $\sum {a_{n}}$ is divergent.

Important: limit approaches 0 means NOTHING

Example 1

$\sum _{n=1}^{\infty }{\sqrt {\frac {n-1}{n}}}$ is divergent by test for divergence:     $\lim _{n\to \infty }a_{n}={\sqrt {1}}\neq 0$

Example 2

Suppose $s_{N}=\sum _{n=1}^{N}{a_{n}}={\sqrt {\frac {N-1}{N}}}$. What do we know about $\sum _{n=1}^{\infty }{a_{n}}$?

$s_{n}=\lim _{N\to \infty }\sum _{n=1}^{N}{a_{n}}=1$; The infinite sum converges to 1.

Note: Harmonic Series

$\sum _{n=1}^{\infty }{\frac {1}{n}}$ is divergent:

${\begin{array}{rcl}\sum _{n=1}^{\infty }{\frac {1}{n}}&=&1+{\frac {1}{2}}+{\frac {1}{3}}+{\frac {1}{4}}+{\frac {1}{5}}+{\frac {1}{6}}+{\frac {1}{7}}+{\frac {1}{8}}+\dots \\&>&1+{\frac {1}{2}}+{\frac {1}{4}}+{\frac {1}{4}}+{\frac {1}{8}}+{\frac {1}{8}}+{\frac {1}{8}}+{\frac {1}{8}}+\dots \\&=&1+{\frac {1}{2}}+{\frac {1}{2}}+{\frac {1}{2}}+\dots \\&\to &\infty \end{array}}$