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'sum of series depends on order'
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| Title of object: |
sum of series depends on order |
| Canonical Name: |
SumOfSeriesDependsOnOrder |
| Type: |
Example |
| Created on: |
2004-11-23 18:28:45 |
| Modified on: |
2006-06-26 13:58:49 |
| Classification: |
msc:40A05, msc:26A06 |
| Keywords: |
conditional convergence |
Revision comment (for changes between this and next version):
Preamble:
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\usepackage{amssymb}
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%\usepackage{psfrag}
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Content:
According to the \PMlinkname{Leibniz' test}{LeibnizEstimateForAlternatingSeries},
the alternating series
$$1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+\frac{1}{7}
-\frac{1}{8}+\frac{1}{9}-\frac{1}{10}+\frac{1}{11}-\frac{1}{12}+-...$$
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
$S = (1-\frac{1}{2})+(\frac{1}{3}-\frac{1}{4})+(\frac{1}{5}-\frac{1}{6})+(\frac{1}{7}
-\frac{1}{8})+(\frac{1}{9}-\frac{1}{10})+...,$
$\frac{1}{2}S = \frac{1}{2}-\frac{1}{4}+\frac{1}{6}-\frac{1}{8}+\frac{1}{10}-+...$
Then we add these two series termwise getting the sum
$1\frac{1}{2}S = 1+\frac{1}{3}-\frac{2}{4}+\frac{1}{5}+\frac{1}{7}
-\frac{2}{8}+\frac{1}{9}+\frac{1}{11}-\frac{2}{12}+...$
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. |
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