Series, Sequences And Convergence Concepts Of Primary .
Series, Sequences and ConvergenceConcepts of primary interest:Convergence: absolute and conditionalTests: ratio, comparison, alternating termRadius of convergenceGeometric, Riemann zeta, harmonic seriesSample calculations:Summing the geometric seriesDemonstration of conditional divergenceApplications of convergence testsConvergence of (1-x)-1 and ezTools of the TradeConverting sums to integralsEstimating remaindersFinite SummationsNf ( x) an ( x x0 )n[SSC.1]n 0Our primary calculus procedures, differentiation and integration, are linear operations. Thederivative of a sum is the sum of derivatives and so on. In the case of a finite sum the operation onthe sum makes sense as long as it makes sense term by term. Infinite sums leave that question open.Is the derivative of the sum the sum of the derivatives taken term-by-term? An infinite sum must beconvergent in order to make any sense. The sum must be strongly and nicely convergent to ensurethat the derivative of an infinite sum is the sum of derivatives taken term-by-term. The tests forconvergence and the types of convergence are discussed below.Infinite Series – a useful example: the Taylor’s series f ( x) an ( x x0 )n[SSC.2]n 0The Taylor’s series with remainder term also plays a role in the discussion of convergence. Itclarifies the conditions required in order that the infinite series can be adequately represented by aContact: tank@alumni.rice.edu
finite number of terms.sn 1 d ff ( x) ss 0 s ! dx 1 d n 1 fwhere Rn ( x, x0 ) n 1 [ n 1]! dx 1 d n 1 f n 1 [ n 1]! dx * (xx (x x )min0n 10 x)x0n 1( x x0 )s Rn ( x, x0 )[SSC.3]for some x * [ x, x0 ] and hence: 1 d n 1 f Rn ( x, x0 ) [ n 1]! dx n 1 (x x )maxn 10[SSC.4]The maximum and minimum are to be found in the interval [x0, x ].The Sweetened Tea Sequence: A certain professor has an 18 oz. cup that he uses to brew tea in amicrowave oven. He observes that the tea is best when it is warm and when it contains between 0.75and 4/3 packets of sweetener per 18 oz. The tea itself is great at one bag per 18 oz and even better if itis stronger. The tea cools more rapidly when the cup is partially filled than when it is fully filled.The professor concludes that he should brew an eighteen-ounce cup with one tea bag and one packetof sweetener and then consume the tea until a fraction a remains. Water is then added up to the 18ounce level, and the tea is re-brewed after tossing in a new teabag and a packet of sweetener. The teabags are removed shortly after the tea is taken from the microwave oven.What is the maximum fraction a of the tea that can remain before the tea is re-brewed each time ifthe beverage is not to exceed the 4/3 packets per 18 ounce sweetener limit?Let the 18 ounces be one unit of volume. Beginning with an empty cup, the first cup brewed has 1packet. A fraction a of that first packet remains when the second cup is brewed with an additionalpacket. The second cup has (1 a) packets. A fraction a of that (1 a) packets remains whenanother packet is added for the third brewing. The third cup contains 1 a (1 a) 1 a a2packets. The pattern is now clear. The nth cup contains 1 a a2 an-1 packets.The finite sum 1 a a2 an-1 is easily evaluated using the algebraic identity:[1 a a2 an-1] [1 – a ] 1 - an[SSC.5]The sweetness of the nth cup is Sn (1 - an)/(1 – a), and the limiting sweetness of the tea is 1/(1 – a) inpackets per eighteen ounces.12/26/2007Handout Series.Tank: Series, Sequences & ConvergenceSSC-2
The problem was to find the maximum value of a such that the sweetness is limited to be less than4/3 packets per eighteen ounces.141/(1 – a) /3 if a /4.Exercise: In an effort to reduce the amount of sweetener consumed, the professor decides to addonly ½ packet of sweetener each brewing. He can tolerate this deprivation as long as he believes thatthe sweetness level will reach 0.75 packets per 18 ounces eventually. How many ounces of each cupshould be consumed before re-brewing if the 0.75 sweetness level is to be reached ‘eventually’?[12 oz.]The Geometric Series (a.k.a. The Sweetened Tea Sequence) The infinite series 1 a a 2 a 3 . a i . a k (1 a ) 1 is the geometric series.k 0The geometric series is absolutely convergent for a 1.Infinite Series and Sequences: An infinite sum of scalar terms such as a iis an infinite series. A sequence is an indexed list ofi 0scalar values such as {b0, b1, b2, . , }. For each infinite series, there is a sequence of particularinterest, the sequence of partial sums.2{ . , Sn , .} {a0, a0 a1, ai , ,i 0n ai 0i, }The sequence has a limit Slim if, for every ε, there exists an integer N such that Slim – Sn ε for alln N. The infinite series converges and has a sum if the sequence of its partial sums has a limit. Itssum is then that limit. When Slim exists, it will be represented as S, the limit of the sequence of partial sums. The remainder after n terms Rn is (S – Sn ) a .i n 1iThe symbol σ represents the sum of a series for which convergence has not been established.Note that σ is not really defined if the series does not converge.12/26/2007Handout Series.Tank: Series, Sequences & ConvergenceSSC-3
The infinite series ai 0i converges only if RN ai N 1i 0 as N .For a series to converge, the sequence of partial sums must approach a limit requiring at minimumthat the terms approach zero as the index grows large.The convergence of one family of infinite series is to be studied before the question of convergenceis developed systematically. The Riemann Zeta function is defined as a series: ( k ).ζ ( p) 1k 1[SSC.6]pIt is the tail-end (large index) terms that must be studied to establish the convergence properties. RN-1 ( k ) . Does this sum approach zero as N becomes very large? The strategy is to sandwich the 1k Npvalue of the summed series between the values of the two integrals.Georg Friedrich Bernhard Riemann (1826 - 1866)(pronounced REE mahn) was a German mathematician who made importantcontributions to analysis and differential geometry, some of them paving theway for the later development of general relativity. Riemann was arguably themost influential mathematician of the middle of the nineteenth century. Hispublished works are a small volume only, but opened up research areascombining analysis with geometry. These would subsequently be major parts ofthe theories of Riemannian geometry, algebraic geometry and complex manifoldtheory. The theory of Riemann surfaces was elaborated by Felix Klein andparticularly Adolf Hurwitz. This area of mathematics was foundational intopology, and in the twenty-first century is still being applied in novel ways tomathematical physics.http://en.wikipedia.org/wiki/G. F. B. RiemannRead the Tools of the Trade section on converting sums to integrals and then carefully prepare asketch to establish that: Ndx xpThe anti-derivatives are: ( 1k k Np) dxN 1 x p 1 1 pdx [1 p ] x( ) xp ln( x)[SSC.7]for p 1for p 1For p 1, both integrals diverge so the sum must diverge.12/26/2007Handout Series.Tank: Series, Sequences & ConvergenceSSC-4
( k ) ln( ) so the sum diverges.For p 1, ln( ) 1k N1 ( p 1) N p 1For p 1,p ( 1k k Np) ( p 1) (1N 1)p 1so both bounds vanish in the large N limit.The series converges.Convergence of the Riemann Zeta Series: converges for p 1 )( k diverges for p 1ζ ( p) 1k 1pThis integral technique is also used to determine bounds on the sum of a series. The first N terms aresummed, and the remainder term is bounded by a pair of integrals. ( 1k Riemann Zeta Series as a name is a little stuffy. In the future, the seriesk 1p) is to be called thep-series. The special case with p 1 is to be called the harmonic series. ( 1 k ) is called the harmonic series. It can be shown to diverge using the integral test leading to The seriesk 1comparison with the function ln(x) . The divergence, however, is very slow. Divergence of the harmonic serieswas first demonstrated by Nicole d'Oresme (ca. 1323-1382), but was mislaid for several centuries (Havil2003, p. 23; Derbyshire 2004, pp. 9-10). The result was proved again by Pietro Mengoli in 1647, by JohannBernoulli in 1687, and by Jakob Bernoulli shortly thereafter (Derbyshire 2004, pp. htmlDivergent Series Nonsense: Begin with the series σ 3n 3 9 27 81 . . One can subtract 3 from this series to findn 1 σ 3 3n 9 27 81 . 3σ suggesting that 2 σ - 3 or σ -3/2. You can generate anyn 2nonsense that you desire by treating a divergent series as meaningful.How convergent is necessary?12/26/2007Handout Series.Tank: Series, Sequences & ConvergenceSSC-5
A series is called conditionally convergent if, the series ak 1k, as written, converges, but the series ak 1kdiverges. A reordering of a conditionally convergent series may converge to a different valueor fail to converge.SEE: Eric W. Weisstein. "Riemann Series Theorem." From MathWorld--A Wolfram Web sTheorem.htmlConsider the series: n 1( 1) n 1 1 1 1 1 1 1 1 . Note that the sum of the positive terms1 2 3 4 5 6 7nappears to be greater than the sum of the absolute values of the negative terms although thestatement is meaningless since both sums diverge. Note that the sum of the negative terms is – (1/2)ζ(1), a series that was demonstrated to diverge by comparison with integrations that bounded theseries sum. The conclusion is that the sum of the positive terms and the sum of the negative termsare divergent. As noted by Boas, the sum can be arranged to sum to any value desired. Pick 1.5. Onesums values until the sum exceeds 1.5 (1 1/3 1/5 1.53) and then adds negative terms until thesum drops below 1.5 (1 1/3 1/5 - 1/2 1.03). One then adds positive values until 1.5 is exceeded.1 1/3 1/5 - 1/2 1/7 1/9 1/11 1/13 1/15 1.52This game can be continued forever as the sum of the positive terms and the sum of the negativeterms separately diverge.1 1/3 1/5 - 1/2 1/7 1/9 1/11 1/13 1/15 -1/4 1.271 1/3 1/5-1/2 1/7 1/9 1/1 1/13 1/15 -1/4 1/17 1/19 1/21 1/23 1/25 1.51Absolute Convergence: A series is absolutely convergent if the series that is the sum of the absolutevalues of each term in the original series is convergent.Rearranging the terms in an absolutely convergent series does not change the sum. n 1 an convergent an absolutely convergentn 1All rearrangements yield the same sum.An Infinite Series as a Function:12/26/2007Handout Series.Tank: Series, Sequences & ConvergenceSSC-6
The Taylor’s series is a representation of a function as an infinite series. If the series converges in aregion, the function is said to be analytic in that region. 1 dn fnn 0 n ! dxf ( x) an ( x x0 ) nn 0( x x0 )n[SSC.8]x0The series is convergent if, at each x, for every ε, there exists an Nε such that1 dn fnn 0 n ! dxNεf ( x) ( x x0 )n ε[SSC.9]x01 dn fIf a single value of Nε can be found such that f ( x) nn 0 n ! dxNε( x x0 )n ε for all x in thex0region, then the series is uniformly convergent in that region. If all the series involved are uniformlyconvergent, then the series for the derivative of f(x) is the series formed by differentiating the seriesfor f(x) term by term, and the series for the integral of f(x) is the series formed by integrating theseries for f(x) term by term. These processes effectively interchange the order in which limits aretaken. Interchanging limiting processes requires uniform convergence to be guaranteed to makesense. This property follows because a uniformly convergent series can effectively be replaced by itsfirst Nε terms and treated as a finite sum.(sometimes: circle of convergence)Radius of Convergence:Each series can be interpreted to be a function of a complex variable. 1 dn fnn 0 n ! dzf ( z ) an ( z z0 ) n 0n( z z0 )n[SSC.10]z0If the function is analytic, then there exists a value R, the radius of convergence, such that thefunction converges at all points z for which z z0 R , at no points for which z z0 R . It mayconverge at some points for which z z0 R and not for others. The value R can be infinity inwhich case the function is an entire function and the series converges everywhere in the complexplane.12/26/2007Handout Series.Tank: Series, Sequences & ConvergenceSSC-7
A function that is analytic in the shaded regionwill have a Taylor’s series expansion about z0that converges at all points inside the circle (ofdivergesoutsideconvergence) of radius R, at no points outsideRthat circle and that it may converge at somezopoints on the circle, but not at others. Theconvergesinsideseries converges uniformly and absolutelyinside any concentric circle with radius R. ( 3)Consider the series xs 0 ( x 3 ) seriesss . It is the geometric series aswhich converges for a 1. Thus thes 0has a radius of convergence of 3 centered on 0. The more general seriess 0 s (b ( z z ))s 00-1has a radius of convergence of b centered on z0. These examples demonstrate thata series with terms that are functions of a variable may converge for a restricted range of values ofthe variable and diverge for values of the variable outside that range. ( Exercise: How does the radius of convergence fors 7The series expansion of (1 x)-1 isx3)s ( x 3 ) compare with that fors?s 0 ( x)n . It follows from the ratio test, that the expansion has an 0radius of convergence of 1 centered on x 0. The sums for n 0 to nmax 50 are compared for x 0.95 and 1.05.Mathematica:converge Table[{n,Sum[(-1) e,PlotRange {0,1}]diverge Table[{n,Sum[(-1) ge,PlotRange All]12/26/2007Handout Series.Tank: Series, Sequences & ConvergenceSSC-8
x 0.95 convergentx 1.05 divergentx 0.99 convergentx 1.01 divergentConvergence Criteria to this Point:A series is just nonsense unless it converges so it is crucial that you learn to test for convergence.What has been learned to this point?Term Behavior: A series cannot converge unless an approaches zero for large n. Geometric Series: 1 a a 2 a 3 . a i . a k (1 a ) 1[SSC.11]k 0-1This series converges to (1 - a) for a 1 as demonstrated by brute force calculation. As thegeometric series is a positive term series, the convergence is absolute. The series diverges for a 1. for p 1( k ) convergesdiverges for p 1Riemann Zeta Series: ζ ( p ) 1k 1p[SSC.12]This family converges for p greater than 1. As they are a positive term series, this convergence is12/26/2007Handout Series.Tank: Series, Sequences & ConvergenceSSC-9
absolute. In the case p 1, the zeta series is the harmonic series which is divergent and stands as theboundary between converging and diverging Riemann zeta series.Tests for ConvergenceThe Sanity Check: A series cannot converge unless an approaches zero for large n. The sanitycheck is not a test for convergence; rather it provides a test to eliminate series that cannot converge.The harmonic series is an example that passes the sanity check, but which fails to converge. That is:passing the sanity check is necessary, but not sufficient.The Integral Test: This test was presented by example in the Riemann zeta introduction. A finitesum of defined values is always convergent. Trap the value of the remainder sum RN of a positiveterm series between two integrals with N dependent limits and show that both integrals approachzero as N becomes large. Thus the remainder term goes to zero. This procedure is a special case ofthe comparison test. The Comparison Test for Convergence: Given that the series to test is ak 1k, find an absolutely bconvergent serieskk 1such that bk ak for all k greater than some N. If such a series can be found, the seriesak is absolutely convergent, and the seriesk 1 ak 1kconverges to the same well-defined value for all re-orderings.The greater than some N appears because convergence is not determined by any finite number of termsin the series; it is determined by the tail-end behavior RN. The Comparison Test for Divergence: Given that the series to test is ak 1k, find a positive term divergent series bk 1ksuch that bk ak for all k greater than some N. If such a series can be found, then the series k 112/26/2007ak diverges.Handout Series.Tank: Series, Sequences & ConvergenceSSC-10
Corollaries of the Comparison Test: i.) If bk 1kis an absolutely convergent series, andanbntends to a finite limit, then the series ak 1kconverges. ii.) If bk 1kis a divergent positive-term series, andanbn tends to a finite limit (non-zero), then the series k 1ak diverges. The Ratio Test: Given that the series to test is ak 1k, form the sequence of the ratios of the nextterm to the current term. 1 absolutely convergentan 1 Limit 1the test fails!n an 1 fails to converge As the test involves the absolute values, the convergent case would be absolutely convergent. Notethat in the case that the limit is 1, the test fails. Some other test must be applied to determine theconvergence or divergence. The ratio test can be motivated by comparison test with the geometricseries in the cases a 1 and 1. In the case of the ratio approaching 1, the test fails. That is: nodefinitive prediction can be made without additional information. One must appeal to morespecialized tests.The ratio test is often applied just after the sanity check.Alternating Term Series: If, after some finite number of terms (for all n greater than some finite N),the terms in a series are alternately positive and negative, an 1 an and the Limit an 0 , the seriesn converges. Note that the restriction an 1 an obviates concerns about reordering the series.More Convergence Tests: The convergence of series is an area of active interest, and there aremany specialized tests for convergence. The discussion of additional tests is to be appended to theTools of the Trade section just before the Problems as time permits or reason dictates.12/26/2007Handout Series.Tank: Series, Sequences & ConvergenceSSC-11
Sample Calculations/Applications of Convergence Tests: 1.) The series 1m 0 11 2m 1 2 m1 2 m 1 can be rewritten as 2 m 1 . This later positive term series dominates the series m 1m 11term by term. As the final ‘p 1’ series diverges, so mustm 1 2 m 1 by the comparisonm 0test.( 1) mis an alternating term series. Clearly am 1 am for all m, andm 0 2 m 1 2.) The series am m 0Limitm am 0 . The series can be summed using a variety of tricks. In the Fourier series handout, the sum isfound to be π4 ( 1) m. m 0 2 m 1 3.) The seriesmm 0Limitm ( 1) mis an alternating term series. Clearly am 1 am for all m, and2m 0 (2 m 1) a am 0 . The series converges to a value defined to be Catalan’s constant (0.915965594 ). 4.) The series am 0m1is an positive term series. Clearly, Limit am 0 .2m m 0 (2 m 1) 11and 1 22m 0 (2 m 1)m 1 (2 m 1) by1 2m 1 (2 m )1 (2 m 1) 14 m 11. The final series, 2m 1 ( m ) 21 ( m)m 1is a positive term series that is dominated term by term2, converges, so all the series converge. 12 π8 ; 2m 1 (2 m 1)1 ( m)m 1 π6225.) Discuss the convergence of the binomial expansion for the inverse of (1 x).(1 x ) 1 s 0set n 1n( n 1)( n 2).( n s 1)s! xs ( 1)( 1 1)( 1 2).( s )s 0s! x s ( 1) s x ss 0It is an alternating term series for positive x. As long as x 0 the terms alternate in sign. For x 1, as 1 as 12/26/2007Handout Series.Tank: Series, Sequences & ConvergenceSSC-12
for all s, and Limit as 0 . The series converges by the alternating terms series test for 0 x 1. Taking as different view, the series is a geometric series with a - x so it converges for - x 1 and hence for x 1.This result subsumes that found using the alternating term approach. This series is an expansion about zero soit must converge for all z such that z R and diverge for all z R. The alternating term series test indicatedthat R must be 1, and the geometric series result confirmed it. Exercise: Replace x by the complex value z. ( 1) s x s s 0 ( z )s.s 0 Let z a i where a is real and 0 a 1. What is ( a i )s? Take the limit a 1. Compare the result withs 0 ( i )½ (1 – i) (1 i)-1. Note than the seriessdoes not converge because an does not approach zero fors 0large n.Mathematica 6: Sum[(- I) ,{s,0,Infinity}] s1/2- /2sMathematica 5.2: Sum @H IL , 8s, 0, Infinity D Sum::div: Sum does not converge. ‚ H Ls s 06.) Discuss the convergence of the binomial expansion for the inverse of (1-x).(1 x ) 1 n( n 1)( n 2).( n s 1)s!s 0set n 1 ( x) s ( 1)( 1 1).( s )s 0s! ( x) s x ss 0The sanity check is valid: Limit as 0 . The term by term ratio is x so, by the ratio test, the series convergess absolutely as long as x 1. Could the ratio test have been applied in the previous example? Note that theratio test result is valid in the complex plane so the series for (1-z)-1, converges everywhere inside a disk ofradius one centered on z 0.z7.) Discuss the convergence of the Taylor’s series expansion of e . ez s 0zss! 1 z z2z3zs . .2 2(3)s!The (s 1) to s term-by-term ratio is /(s 1). The nth term in the sum has index s n – 1.zThe ratio test is:12/26/2007Handout Series.Tank: Series, Sequences & ConvergenceSSC-13
1 convergentan 1az Limit 1test fails!Here, Limit n 1 Limit 0 for z finiten n s anasn 1 divergent n so, by the ratio test, the series converges absolutely for any finite z . A function with an unbounded or infiniteradius of convergence is well-defined and analytic over the entire complex plane and is called an entirefunction.-18.) Discuss the convergence of the Taylor’s series expansion of [1 3x ] . [1 3x] 1 ( 3 x ) 1 3 x 9 x 2 27 x3 .ss 0n 1 3 xThe absolute value of the term-by-term ratio is ( )( 3 x )n 3 x 3 x . The ratio test is: 1 convergentan 1 Limit Limit 3 x 3 x 1 test fails ! so, by the ratio test, the series convergesn n an 1 divergent 1absolutely for x /3.Tools of the Trade:Converting Sums to IntegralsIt is said that an integral is a sum of a huge number of small contributions, but some precision isrequired before the statement becomes useful. Beginning with a function f(t) and a sequence ofi Nvalues for t {t1,t2,t3, .,tN}, the sum f (t ) does not represent the integral i 1it t f (t ) dt even if agreat many closely spaced values of t are used. Nothing has been included in the sum to representi Ndt. One requires f (t ) Δti 1ii( 2 ) [twhere Δti 1i 1 ti 1 ] is the average interval betweeni Nsequential values of t values at ti. For well-behaved cases, the expression f (t ) Δti 1iiapproaches theRiemann sum definition of an integral as the t-axis is chopped up more and more finely. As12/26/2007Handout Series.Tank: Series, Sequences & ConvergenceSSC-14
illustrated below, in the limit Δt goes to zero, the sumi N f (t ) Δtii 1curve between t and t . That is: it represents t t iapproaches the area under thef (t ) dt provided the sequence of sums converges,and life is good. The theory of integration is not the topic of this passage. The goal is simply toremind you that the Δt must be factored out of each term that is being summed in order to identifythe integrand.f(t N)f(t)f(ti)f(t 1)f(tk )ΔtΔttt1 t2tiarea f(t k ) Δtt tktNt In the Fourier series handout discussion of the inner product, the function h(t) g(t) f(t) wasconsidered at N equally spaced points between –T/2 and T/2. This leads to the summ N h(tm 1( 2 ) (T 2 N ) m (T N ) for (1 m N) have equal spacingthe N points tm TΔt TNm) whereand arecentered in each Δt wide interval. As the number of terms gets large, the sum must be divided by Nm Nto ensure a result finite. This leads to h(t ) ( 1 N ) .mm 1requires that Δt Tm NNbe explicitly factored from each term in the sum. Thusm N h(t ) ( 1 N ) [( 1T ) h(tm 1mThe rule for converting sums to integralm 1mm N( N ) [( 1T ) h(t)] Tm 1m)] Δt which becomes T / 2 T / 2( ) 1 f (t ) dt T as N gets large and Δt small. Converting alternating term sums: m 012/26/2007( 1)m2m 1 π4 0.7853982Handout Series.Tank: Series, Sequences & ConvergenceSSC-15
Anchor the evaluation. Sum through m 7:( 1)7 m 0.75426802m 1m 0( 1) The goal is to first estimate and then to provide bounds for the value ofm2m 1m 8byconverting the remainder sum to an integral.Converting a sum to an integral is sure to fail unless the change in the value of terms from one termto the next is small compared to the absolute value of the terms. The conversion may fail anyway,but unless this condition is met, it’s hopeless. The series above is an alternating (sign) term series.The magnitude of the change in value of the sum from term to term is twice the magnitude of theterms. A trick is needed. Combine adjacent positive and negative terms in pairs. It is a trick; thinkabout the sums below to verify that they are equal.( 1) m2m 1m 8 1 1 4 1 4 3 4 4 1622 16 3 The last sum has the advantage that it is a sum of positive terms that vanish more rapidly than do thealternating terms in the original series. It has better convergence properties.To convert to an integral, Δ must be factored from each term, buttakes on every integer greaterthan 4, the starting point, so Δ 1 . 4 162 2 16 3 4 167 m 0 2 Δ 2 16 3 164( 1) m 2m 14162221 19 d ln 0.278064 16 34 17 2d 0.7821 16 3The exact value is π/4 0.785398. With patience one can do better. Exercise: Show that: 4 N 321 1 1 N 16 2 16 3 d N 4 1 4 3 d 4 N 1 u du 4 . Use the result to()evaluate the integral above.Bounding the sum: For suitable choices of limits, the integrals can provide both upper and lower12/26/2007Handout Series.Tank: Series, Sequences & ConvergenceSSC-16
bounds for the sum. A careful sketch shows that for a monotone decreasing function F(n), N F ( x) dx F (n) N 1n NF ( x) dx .The area of the shaded region is F(N), and F(x) asdrawn is a monotone decreasing function. It followsthat:N 1 NF ( x) dx F ( N ) N 1NF ( x ) dxExercise: Argue that, for a monotone decreasingfunction, N F ( x) dx F (n) N 1n NF ( x) dxThe shaded area is F(N) times 1. Make a sketch toshow that N 1F ( x) dx F ( N ) NN 1NF ( x ) dx giventhat F(x) is monotone decreasing.7 m 0( 1) m 2 d 2m 116 2 16 3m 04or equivalently0.7821 m 0( 1)( 1)m2m 1( 1)7 m 0m2m 1 31622d 16 3m2m 1 0.7900. These results are consistent with a sum of π/4 0.7854. A current estimate might be chosen to be:7 m 0( 1) m2m 1 163.522 1 17 d 0.7542680 ln 0.7855587 16 3 4 15 It seems that splitting the difference is reasonable so long as one remembers the bounds on theproven range.Expansion/Identity SummaryDouble Factorial: n!! n(n-2)(n-4) {terminating with 2 or 1.}Function12/26/2007SeriesHandout Series.Tank: Series, Sequences & ConvergenceSummationRconvSSC-17
f (x) f (x0 ) (1)f(x)(x y)ndfdx22 1 d fx x ()x 2 x ( x x ) .0sin(x)cos(x)1 ln[1 x](1 x)(1 x)eernxixs-n ( n ns!)!s! x n s y sn-x 2 n 1( 1) ( 2 n 1)!n 0 n ( 1)nn 0 x2n( 2 n )! ½π ( 1)If n is a positive integer, the series terminates.23 1 s 11xs1finite sum no convergence issue xn n 0 n !xxx1 x 2 3! 4! ( 1)x 3 1i3i x 5 1i3i5i x 7 6 2i4i5 2i4i6i7x 1 11 3 5 2 x 3x 5 xx2 x4 x6 2! 4! 6!x 3 x5 x 7x 3! 5! 7! x2 mx 2 m 1 i ( 2 m ) ! ( 2 n 1) ! mm 0π1 xnnr ( r 1)( r 2).( r s 1)s!4x2x3 x 4x51 ix i i 23! 4!4!n 1n 1r ( r 1)r ( r 1)( r 2) 31 r x 2! x 2 x .3!Arctan(x)tanh(x)( x x0 )s 0x 3 2 x5 17 x 7 62 x 7 3 15 315 2835x 2 x3 x 4 x5 x 6x 2 3 4 5 6x3 x5 x 7x 3 5 7sinh(x)( x y)x0x x cosh(x)0x2 x4 x6 2! 4! 6!Arcsin(x)Arctan(1/x) 2! dx0s 1 d f ss 0 s ! dxxny0 n xn-1y1 n(n-1)/2! xn-2y2 n(n-1) (n-2)/3! xn-3y3 .x3 x5 x 7x 3! 5! 7!tan(x)0 x3 (2m 1)!! x 2 m 1 6 m 2 (2m)!!(2m 1)1( 1)( 2 n 1) x 2 n 11 ( 1)n 02 n 0x2n( 2 n )! x 2 n 1( 2 n 1)! n 0 n 0nn 1x 2 n 1( 2 n 1) π x 3 2 x 5 17 x 7 62 x 7x 3 15 315 2835Taylor’s Series expansion of f(x) about the argument value xo:12/26/2007Handout Series.Tank: Series, Sequences & ConvergenceSSC-18
f ( x) f ( x0 ) (1)dfdxx01 d2 f( x x ) 20 2! dxx0 1 ds f2( x x ) . s0s 0 s ! dxx0( x x0 )sBinomial Theorem: The theorem is an algebraic identity used to compute positive integer powers ofa binomial.( x y)n( )n β n , s x n s y s ; β n , s ns s 0n!n(n 1)(n 2).(n s 1) s!( n s )! s !TO APPLY, convert to the form (1 x)n where x 1 and use the form:(1 x)n 1 nx n( n 1) 2 n(n 1)( n 2) 3n( n 1)(n 2).( n s 1) sx x . x2!3!s!Use the convention that only positive square roots considered for forms likerepresent distances and thata 2 z 2 as thez 2 z ; (r z ) 2 r z ;. unless you have a reason to dootherwise.Euler's Identity: a relation between the exponential and the sinusoidal functionsalso known as: The Euler-Lagrange identityeiθ cos θ i sin θProblems1.) Express the following series in summation form. Interpret the as it just keeps on going.1 1 1 1a.) 1 .3 5 7 9c.) 1 x2x4x6 2 2 3 4 6!b.)1 1 1 1 1 .4 9 16 25 36d.) x x3x5x7 2 3 2 3 4
Converting sums to integrals Estimating remainders Finite Summations (0 0 N n n n f xaxx) [SSC.1] Our primary calculus procedures, differentiation and integration, are linear operations. The derivative of a sum is the sum of derivatives and so on. In the case of a finite sum the operation on the sum makes sense as long as it makes .
Sequences and their convergence, convergence and divergence of infinite series. Geometric series. P-series. A necessary condition for convergence. Comparison test, ratio test, absolute & conditional convergence of alternating series. Expansion of functions: Macluarins & Taylor's expansion with remainder form. Indeterminate forms. L .
Convergence of sequence and series-tests for convergence: Integral Test, Comparison test, D’Alembert’s Ratio test, Cauchy root test, Raabe’s Test, Leibnitz test for alternate series. Power series: Radius of convergence and Interval of convergence, Taylor's series, series for exponent
Of the three closely associated convergences—technological convergence, media convergence, and network convergence—consumers most often directly engage with technological convergence. Technological convergent devices share three key characteristics. First, converged devices can execute multiple functions to serve blended purpose.
In Unconditional Convergence (2011), Dani Rodrik documented that manufacturing industries exhibit unconditional convergence in labor productivity. We provide a novel semi-parametric specification for convergence equations and show that the pace of convergence varies systematically with country-specific characteristics.
Some Applications of the Bounded Convergence Theorem for an Introductory Course in Analysis JONATHAN W. LEWIN Kennesaw College, Marietta, GA 30061 The Arzela bounded convergence theorem is the special case of the Lebesgue dominated convergence theorem in which the functions are assumed to be Riemann integrable. THE BOUNDED CONVERGENCE THEOREM.
Convergence of Sequence for SAGA/Prox-SVRG As-suming only convexity, we prove the almost sure global convergence of the sequences generated by SAGA (Theo-rem2.1) and Prox-SVRG with "Option I" (Theorem2.2), which is new to the literature. Moreover, for Prox-SVRG with "Option I", an O(1 k) ergodic convergence rate for
sequences (DNA, RNA, or amino acid sequences), high sequence similarity usually implies signi cant functional or structural similarity." D. Gus eld, Algorithms on strings, trees and sequences Note that the converse is not true: \ . similar sequences yield similar structures, but quite di erent sequences can produce remarkably similar structures."
Complex sequences and series An infinite sequence of complex numbers, denoted by {zn}, can be considered as a function defined on a set of positive integers into the unextended complex plane. For example, we take zn n 1 2n so that the complex sequence is {zn} ˆ1 i 2, 2 i 22, 3 i 23,··· . Convergence of complex sequences
