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BMCC MTH 253 - Infinite Sequences and Series

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MTH 253 Calculus (Other Topics)Does the Series Converge?The Harmonic SeriesSlide 4The Integral TestSlide 6The Integral Test: Examplesp-SeriesSlide 9Slide 10MTH 253Calculus (Other Topics)Chapter 11 – Infinite Sequences and SeriesSection 11.3 – The Integral TestCopyright © 2009 by Ron Wallace, all rights reserved.Does the Series Converge?1kka�=�10 Tests for ConvergenceGeometric SeriesN-th Term Test (Divergence Test)Integral Testp-Series TestComparison TestLimit Comparison TestRatio TestRoot TestAlternating Series TestAbsolute Convergence TestEach test has it limitations (i.e. conditions where the test fails).The test tells you nothing!The Harmonic Series11 1 1 11 ...2 3 4kk�== + + + +�12 13 2111213 1n nss ss ss sn-== += +���= +{ }1nns�=is a strictly increasing sequenceConverges if there is an upper bound.The Harmonic Series11 1 1 11 ...2 3 4kk�== + + + +�224 2 2221 1 1 212 2 2 21 1 1 1 33 4 4 4 2 12nss s s sns= + > + == = + + > + + =���+>Diverges because there is no possible upper bound.12nM+>Since for any possible upper bound M, you can find an n where …The Integral TestNOTE:For all series in this section, it will be assumed that each term is non-negative.That is, given thenka�0, ka k� "The Integral TestLet f(x) be a decreasing continuous function over [N,) such that f(k)=ak for all k  N.ka�1 2 3 54 76a3a2a1...f(x)1 2 3 54 76a3a2a1...f(x)a2a3a4a5a6a7a1a2a3a4a5a6 1( )kNk Nf x dx a��= +��� If the Integral converges, so does the series.Upper bound for an increasing sequence of partial sums. ( )kNk Nf x dx a��=��� If the Integral diverges, so does the series.NO upper bound for an increasing sequence of partial sums.The Integral Test: Examples21 kkdxxx 1 1 22 2 ln21 121udxu23)24(1kdxx )24(1 1 23611 121621 6 23udxu Divergent! Convergent!When does this test fail? When it can’t be integrated?p-Seriespk1p=1p=2p=1/2...3121111kThe divergent harmonic series....91411112k...3121111121kkp-Seriespk1For p  0Divergent (nth-term test) if p  01p qpk kk-= =� � �Note: -p = q  0limqkk��=�p-Seriespk1 11px�=�For p ≠ 1 and p > 0ppbpxxpbpp111lim1111 1 Convergent if p > 1Divergent if 0 < p 


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