Difference between revisions of "Harmonic series"

(Big revision of article)
m (proofreading)
Line 6: Line 6:
 
   
 
   
 
The alternating harmonic series,
 
The alternating harmonic series,
<math>\displaystyle\sum_{i=1}^{\infty}\frac{(-1)^{i+1}}{i}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots</math> , though approaches <math> \ln 2</math>.
+
<math>\displaystyle\sum_{i=1}^{\infty}\frac{(-1)^{i+1}}{i}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots</math> , though, approaches <math> \ln 2</math>.
  
The general harmonic series, <math>\displaystyle\sum_{i=1}^{\infty}\frac{1}{ai +b}</math> has its value depending on the value of the constants <math>a</math> and <math>b</math>.
+
The general harmonic series, <math>\displaystyle\sum_{i=1}^{\infty}\frac{1}{ai +b}</math>, has its value depending on the value of the constants <math>a</math> and <math>b</math>.
  
 
The [[zeta-function]] is a harmonic series when the input is one.
 
The [[zeta-function]] is a harmonic series when the input is one.
  
 
== How to solve ==
 
== How to solve ==

Revision as of 12:59, 28 June 2006

There are several types of harmonic series.

The the most basic harmonic series is the infinite sum $\displaystyle\sum_{i=1}^{\infty}\frac{1}{i}=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots$ This sum slowly approaches infinity.

The alternating harmonic series, $\displaystyle\sum_{i=1}^{\infty}\frac{(-1)^{i+1}}{i}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots$ , though, approaches $\ln 2$.

The general harmonic series, $\displaystyle\sum_{i=1}^{\infty}\frac{1}{ai +b}$, has its value depending on the value of the constants $a$ and $b$.

The zeta-function is a harmonic series when the input is one.

How to solve

Invalid username
Login to AoPS