Difference between revisions of "Harmonic series"

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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>.
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<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> \displaystyle \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>.

Revision as of 20:18, 22 August 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 $\displaystyle \ln 2$.

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