sum of series depends on orderSumOfSeriesDependsOnOrder2004-11-23 18:28:452009-01-03 18:45:57Examplemanipulating convergent seriesconditional convergence% this is the default PlanetMath preamble. as your knowledge
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% define commands hereAccording to the \PMlinkname{Leibniz' test}{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 \PMlinkname{natural logarithm}{NaturalLogarithm2}).\, Denote it by $S$.\, We can \PMlinkescapetext{group pairwise its terms and multiply each term} 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
$1\frac{1}{2}S = 1+\frac{1}{3}-\frac{2}{4}+\frac{1}{5}+\frac{1}{7}
-\frac{2}{8}+\frac{1}{9}+\frac{1}{11}-\frac{2}{12}+\ldots.$
Hence, this last series \PMlinkescapetext{contains} exactly the same \PMlinkescapetext{terms} 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 \PMlinkescapetext{terms} is the divergent harmonic series.\\
P. S.\; -- For justification of the used manipulations of the series, see the \PMlinkescapetext{parent} entry.