# sum of series depends on order

According to the Leibniz’ test (http://planetmath.org/LeibnizEstimateForAlternatingSeries), the alternating series  $1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+\frac{1}{7}-% \frac{1}{8}+\frac{1}{9}-\frac{1}{10}+\frac{1}{11}-\frac{1}{12}+-\ldots$

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

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

Then we add these two series termwise getting the sum

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

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