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InfiniteSeriesSynopsis
(2008-11-16,
DickFurnas
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---+ !!Infinite Series: A Synopsis %BEGINOVERVIEW% %ENDOVERVIEW% ---+ Famous Series %INCLUDE{FamousSeries}% ---+ [[Useful Limits]] %INCLUDE{UsefulLimits}% ---+ Useful Inequalities %INCLUDE{UsefulInequalities}% ---+ New Series From Old %INCLUDE{NewSeriesFromOld}% ---+ [[Convergence Tests]] %INCLUDE{ConvergenceTests]]% ---+ Tips for Series * Often the hardest part of showing convergence or divergence of a series is the indecision: _What do I believe it does?_ After all, you'll have a tough time showing a series converges if it doesn't! * The limits listed in [[UsefulLimits]] can help a lot with the _Test for Divergence._ Together with inequalities you can often get an idea of what to try to show. If the individual terms of the series "look like" <latex>\frac{n^3}{n^4}</latex> as <latex>n \rightarrow \infty</latex> then the series "looks like" <latex>\frac{1}{n}</latex> and you will want to show it diverges, perhaps even setting up a comparison, or limit comparison with 1/n itself. * Many limits boil down to "look like" ratios of polynomials after stripping out trig functions using the [[Useful Inequalities]] for trig functions. * The eventual behavior that <latex>\ln{n} < n^k </latex> for any <latex>k > 0 </latex> leads to the peculiar rule of thumb that in lots of ratios _ln(n)_ "looks like" 1 since any positive power of _n_ will dominate it: * informally, <latex>\sum \frac{\ln{n}}{n^2}</latex> "looks like" <latex>\sum\frac{1}{n^2}</latex> so converges * more carefully, <latex>\ln{n} < \sqrt{n}</latex> (eventually), %BR%so<latex>\sum\frac{\ln{n}}{n^2} < \frac{\sqrt{n}}{n^2} = \sum\frac{1}{n^{\frac{3}{2}}}</latex>, %BR%which is a convergent p-series. --- -- Main.DickFurnas - 21 Oct 2008
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Topic revision: r7 - 2008-11-16 - 22:51:58 -
DickFurnas
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