# sum of series depends on order

According to the Leibniz’ test (http://planetmath.org/LeibnizEstimateForAlternatingSeries),
the alternating series^{}

$$1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+\frac{1}{7}-\frac{1}{8}+\frac{1}{9}-\frac{1}{10}+\frac{1}{11}-\frac{1}{12}+-\mathrm{\dots}$$ |

is convergent^{} and has a positive sum ($=\mathrm{ln}2$; see the natural logarithm^{} (http://planetmath.org/NaturalLogarithm2)). Denote it by $S$. We can by $\frac{1}{2}$ getting the two series
$S=(1-\frac{1}{2})+(\frac{1}{3}-\frac{1}{4})+(\frac{1}{5}-\frac{1}{6})+(\frac{1}{7}-\frac{1}{8})+(\frac{1}{9}-\frac{1}{10})+\mathrm{\dots},$

$\frac{1}{2}S=\frac{1}{2}-\frac{1}{4}+\frac{1}{6}-\frac{1}{8}+\frac{1}{10}-+\mathrm{\dots}.$

Then we add these two series termwise getting the sum

$1\u2064\frac{1}{2}S=1+\frac{1}{3}-\frac{2}{4}+\frac{1}{5}+\frac{1}{7}-\frac{2}{8}+\frac{1}{9}+\frac{1}{11}-\frac{2}{12}+\mathrm{\dots}.$

Hence, this last series exactly the same as the original, but its sum is fifty percent greater. This is possible because the original series is not absolutely convergent: the series which is formed of the absolute values^{} of its is the divergent harmonic series^{}.

P. S. – For justification of the used manipulations of the series, see the entry.

Title | sum of series depends on order |

Canonical name | SumOfSeriesDependsOnOrder |

Date of creation | 2013-03-22 14:50:59 |

Last modified on | 2013-03-22 14:50:59 |

Owner | pahio (2872) |

Last modified by | pahio (2872) |

Numerical id | 16 |

Author | pahio (2872) |

Entry type | Example |

Classification | msc 26A06 |

Classification | msc 40A05 |

Related topic | AbsoluteConvergence |

Related topic | OrderOfFactorsInInfiniteProduct |

Related topic | AlternatingHarmonicSeries |

Related topic | ConditionallyConvergentSeries |

Related topic | ConvergingAlternatingSeriesNotSatisfyingAllLeibnizConditions |

Related topic | FiniteChangesInConvergentSeries |

Related topic | FiniteChangesInConvergentSeries2 |