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Home , Convergence tests, Divergent series, Riemann series theorem

, and Convergence

Concepts of primary interest: Convergence: absolute and conditional Tests: ratio, comparison, alternating term Geometric, zeta, harmonic series

Sample calculations: Summing the Demonstration of conditional Applications of convergence tests Convergence of (1-x)-1 and ez Tools of the Trade Converting sums to Estimating remainders

Finite Summations

N n f ()xaxx=−∑ n (0 ) [SSC.1] n=0 Our primary procedures, differentiation and integration, are linear operations. The of a sum is the sum of and so on. In the case of a finite sum the operation on the 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 be convergent in order to make any sense. The sum must be strongly and nicely convergent to ensure that the derivative of an infinite sum is the sum of derivatives taken term-by-term. The tests for convergence and the types of convergence are discussed below.

Infinite Series – a useful example: the Taylor’s series

∞ n f ()xaxx=−∑ n (0 ) [SSC.2] n=0 The Taylor’s series with remainder term also plays a role in the discussion of convergence. It clarifies the conditions required in order that the infinite series can be adequately represented by a

Contact: [email protected] finite number of terms.

n s ⎛⎞1 df s f ()xx=−x+R(,xx ) [SSC.3] ∑⎜⎟ s x0 ()0n 0 s=0 ⎝⎠sdx!

n+1 ⎛⎞1 df n+1 where R (,xx )=−x xforsome* x∈ xx , and hence: n 00⎜⎟n+1 |x* () [0] ⎝⎠[]ndx+1!

nn++11 ⎛⎞11df nn+11⎛⎞df + xx−≤ Rxx(, ) ≤ xx− [SSC.4] ⎜⎟nn++11||min() 0 n 0 ⎜⎟ max() 0 ⎝⎠[]ndx++1! ⎝⎠[]ndx1!

The maximum and minimum are to be found in the interval [x0, x ].

The Sweetened Tea : A certain professor has an 18 oz. cup that he uses to brew tea in a microwave oven. He observes that the tea is best when it is warm and when it contains between 0.75 4 and /3 packets of sweetener per 18 oz. The tea itself is great at one bag per 18 oz and even better if it is 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 packet of sweetener and then consume the tea until a fraction a remains. Water is then added up to the 18- ounce level, and the tea is re-brewed after tossing in a new teabag and a packet of sweetener. The tea bags 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 if 4 the beverage is not to exceed the /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 1 packet. A fraction a of that first packet remains when the second cup is brewed with an additional packet. The second cup has (1 + a) packets. A fraction a of that (1 + a) packets remains when another packet is added for the third brewing. The third cup contains 1 + a (1 + a) = 1 + a + a2 th 2 n-1 packets. The pattern is now clear. The n cup contains 1 + a + a + … + a packets.

2 n-1 The finite sum 1 + a + a + … + a is easily evaluated using the algebraic identity: 2 n-1 n [1 + a + a + … + a ] [1 – a ] = 1 - a [SSC.5]

th n 1 The sweetness of the n cup is Sn = (1 - a )/(1 – a), and the limiting sweetness of the tea is /(1 – a) in packets per eighteen ounces.

12/26/2007 Handout Series.Tank: Series, Sequences & Convergence SSC-2

The problem was to find the maximum value of a such that the sweetness is limited to be less than 4 /3 packets per eighteen ounces.

1 4 1 /(1 – a) < /3 if a < /4.

Exercise: In an effort to reduce the amount of sweetener consumed, the professor decides to add only ½ packet of sweetener each brewing. He can tolerate this deprivation as long as he believes that the sweetness level will reach 0.75 packets per 18 ounces eventually. How many ounces of each cup should 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++aa23 + a ++... aik + ... =∑ a = (1 − a )−1is the geometric series. k =0 The geometric series is absolutely convergent for |a| < 1.

Infinite Series and Sequences:

∞ An infinite sum of scalar terms such as ∑a−i is an infinite series. A sequence is an indexed list of i=0

scalar values such as {b0, b1, b2, …. , }. For each infinite series, there is a sequence of particular interest, the sequence of partial sums.

2 n { …. , Sn , ….} = {a0, a0 + a1, ∑ai , … , ∑ai , … } i=0 i=0

The sequence has a limit Slim if, for every ε, there exists an integer N such that | Slim – Sn| < ε for all n > N. The infinite series converges and has a sum if the sequence of its partial sums has a limit. Its sum 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 ) = ∑ ai . in=+1 The 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.

12/26/2007 Handout Series.Tank: Series, Sequences & Convergence SSC-3 ∞ ∞ The infinite series ∑ai converges only if RN = ∑ aNi →→0as ∞. i=0 iN=+1

For a series to converge, the sequence of partial sums must approach a limit requiring at minimum that 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 convergence is developed systematically. The is defined as a series:

∞ ζ ()p = 1 . [SSC.6] ∑( p ) k =1 k

It is the tail-end (large index) terms that must be studied to establish the convergence properties. RN-1

∞ = 1 . Does this sum approach zero as N becomes very large? The strategy is to sandwich the ∑ ( p ) kN= k value of the summed series between the values of the two integrals.

Georg Friedrich (1826 - 1866)

(pronounced REE mahn) was a German mathematician who made important contributions to analysis and differential geometry, some of them paving the way for the later development of general relativity. Riemann was arguably the most influential mathematician of the middle of the nineteenth century. His published works are a small volume only, but opened up research areas combining analysis with geometry. These would subsequently be major parts of the theories of , algebraic geometry and complex manifold theory. The theory of Riemann surfaces was elaborated by Felix Klein and particularly Adolf Hurwitz. This area of was foundational in topology, and in the twenty-first century is still being applied in novel ways to mathematical physics. http://en.wikipedia.org/wiki/G._F._B._Riemann

Read the Tools of the Trade section on converting sums to integrals and then carefully prepare a sketch to establish that:

∞ ∞dx 1 ∞dx <

−1 dx ⎪⎧ 1f−≠px()1− p or p1 The anti-derivatives are: → [] ∫ p ⎨ x ⎩⎪ ln(xp ) for= 1 For p < 1, both integrals diverge so the sum must diverge.

12/26/2007 Handout Series.Tank: Series, Sequences & Convergence SSC-4 ∞ For p = 1, ln()∞<1 1, <<1 so both bounds vanish in the large N limit. p−1 ∑ ( k p ) p−1 (1)pN− kN= (1)1pN−−() The series converges. Convergence of the Riemann Zeta Series: ∞ ⎧convergesforp > 1 ζ ()p =⇒1 ∑( k p ) ⎨ k=1 ⎩ diverges forp ≤ 1

This technique is also used to determine bounds on the sum of a series. The first N terms are summed, and the remainder term is bounded by a pair of integrals.

∞ Riemann Zeta Series as a name is a little stuffy. In the future, the series 1 is to be called the ∑( p ) k =1 k p-series. The special case with p = 1 is to be called the harmonic series.

∞ 1 The series ∑(k )is called the harmonic series. It can be shown to diverge using the integral test leading to k=1 comparison with the function ln(x) . The divergence, however, is very slow. Divergence of the harmonic series was first demonstrated by Nicole d'Oresme (ca. 1323-1382), but was mislaid for several centuries (Havil 2003, p. 23; Derbyshire 2004, pp. 9-10). The result was proved again by Pietro Mengoli in 1647, by Johann Bernoulli in 1687, and by Jakob Bernoulli shortly thereafter (Derbyshire 2004, pp. 9-10). http://mathworld.wolfram.com/HarmonicSeries.html

Divergent Series = Nonsense:

∞ Begin with the series σ ==++++∑ 3n 3 9 27 81 ... . One can subtract 3 from this series to find n=1

∞ n 3 σ −=3∑ 3 =+ 9 27 + 81 + ... = 3σ suggesting that 2 σ = - 3 or σ = - /2. You can generate any n=2 nonsense that you desire by treating a as meaningful.

How convergent is necessary?

12/26/2007 Handout Series.Tank: Series, Sequences & Convergence SSC-5 ∞ A series is called conditionally convergent if, the series ∑ak , as written, converges, but the series k=1

∞ ∑ ak diverges. A reordering of a conditionally may converge to a different value k=1 or fail to converge.

SEE: Eric W. Weisstein. "." From MathWorld--A Wolfram Web Resource. http://mathworld.wolfram.com/RiemannSeriesTheorem.html

∞ (1)− n+1 1 1 1 1 1 1 1 Consider the series: ∑ =− +−+−+−. Note that the sum of the positive terms n=1 n 1234567 appears to be greater than the sum of the absolute values of the negative terms although the 1 statement is meaningless since both sums diverge. Note that the sum of the negative terms is – ( /2) ζ(1), a series that was demonstrated to diverge by comparison with integrations that bounded the series sum. The conclusion is that the sum of the positive terms and the sum of the negative terms are divergent. As noted by Boas, the sum can be arranged to sum to any value desired. Pick 1.5. One 1 1 sums values until the sum exceeds 1.5 (1 + /3 + /5 ≈ 1.53) and then adds negative terms until the 1 1 1 sum drops below 1.5 (1 + /3 + /5 - /2 ≈ 1.03). One then adds positive values until 1.5 is exceeded. 1 1 1 1 1 1 1 1 1 + /3 + /5 - /2 + /7 + /9 + /11 + /13 + /15 ≈ 1.52 This game can be continued forever as the sum of the positive terms and the sum of the negative terms separately diverge. 1 1 1 1 1 1 1 1 1 1 + /3 + /5 - /2 + /7 + /9 + /11 + /13 + /15 - /4 ≈1.27 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1+ /3+ /5- /2 + /7+ /9 + /1 + /13+ /15 - /4+ /17 + /19 + /21 + /23 + /25 ≈1.51

Absolute Convergence: A series is absolutely convergent if the series that is the sum of the absolute values of each term in the original series is convergent. Rearranging the terms in an absolutely convergent series does not change the sum.

∞∞ ∑∑aannconvergent⇒ absolutely convergent nn==11 All rearrangements yield the same sum.

An Infinite Series as a Function:

12/26/2007 Handout Series.Tank: Series, Sequences & Convergence SSC-6 The Taylor’s series is a representation of a function as an infinite series. If the series converges in a region, the function is said to be analytic in that region.

∞∞n nn1 df f ()xaxx xx =−=∑∑n ()00n ( −) [SSC.8] nn==00ndx! x0

The series is convergent if, at each x, for every ε, there exists an Nε such that

Nε n 1 df n fx() x x [SSC.9] −−∑ n ()0 <ε n=0 ndx! x0

Nε n 1 df n If a single value of N can be found such that fx()− x−< x ε for all x in the ε ∑ n ()0 n=0 ndx! x0 region, then the series is uniformly convergent in that region. If all the series involved are uniformly convergent, then the series for the derivative of f(x) is the series formed by differentiating the series for f(x) term by term, and the series for the integral of f(x) is the series formed by integrating the series for f(x) term by term. These processes effectively interchange the order in which limits are taken. Interchanging limiting processes requires to be guaranteed to make sense. This property follows because a uniformly convergent series can effectively be replaced by its

first Nε terms and treated as a finite sum.

Radius of Convergence: (sometimes: circle of convergence) Each series can be interpreted to be a function of a complex variable.

∞∞n n1 df n f ()z azz zz =−=∑∑n ()0n ( −0) [SSC.10] nn==00ndz! z0 If the function is analytic, then there exists a value R, the radius of convergence, such that the

function converges at all points z for which zz− 0 < R, at no points for which zz−>0 R. It may converge at some points for which zz− 0 = R and not for others. The value R can be in which case the function is an entire function and the series converges everywhere in the complex plane.

12/26/2007 Handout Series.Tank: Series, Sequences & Convergence SSC-7

A function that is analytic in the shaded region

will have a Taylor’s series expansion about z0 that converges at all points inside the circle (of diverges convergence) of radius R, at no points outside outside R that circle and that it may converge at some zo converges points on the circle, but not at others. The inside series converges uniformly and absolutely inside any concentric circle with radius < R.

∞ s ∞ Consider the series x . It is the geometric series as which converges for |a| < 1. Thus the ∑( 3) ∑ s=0 s=0

∞ s series x has a radius of convergence of 3 centered on 0. The more general series ∑( 3) s=0

∞ s -1 ∑(bz()− z0 ) has a radius of convergence of b centered on z0. These examples demonstrate that s=0 a series with terms that are functions of a variable may converge for a restricted range of values of the variable and diverge for values of the variable outside that range.

∞ s ∞ s Exercise: How does the radius of convergence for x compare with that for x ? ∑( 3) ∑( 3) s=7 s=0

∞ The series expansion of (1 + x)-1 is ∑()−x n . It follows from the , that the expansion has a n=0

radius 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)^m (95/100)m ,{m,0,n}]},{n,0,50}] ListLinePlot[converge,PlotRange → {0,1}] diverge = Table[{n,Sum[(-1)^m (105/100)m ,{m,0,n}]},{n,0,50}]; ListLinePlot[diverge,PlotRange → All]

12/26/2007 Handout Series.Tank: Series, Sequences & Convergence SSC-8

x = 0.95 convergent x = 1.05 divergent

x = 0.99 convergent x = 1.01 divergent

Convergence 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++aa23 + a ++... aik + ... =∑ a = (1 − a )−1 [SSC.11] k=0 This series converges to (1 - a)-1 for |a| < 1 as demonstrated by brute force calculation. As the geometric series is a positive term series, the convergence is absolute. The series diverges for |a| > 1.

∞ ⎧convergesforp > 1 Riemann Zeta Series: ζ ()p =⇒1 [SSC.12] ∑( k p ) ⎨ k=1 ⎩ diverges forp ≤ 1 This family converges for p greater than 1. As they are a positive term series, this convergence is

12/26/2007 Handout Series.Tank: Series, Sequences & Convergence SSC-9 absolute. In the case p =1, the zeta series is the harmonic series which is divergent and stands as the boundary between converging and diverging Riemann zeta series.

Tests for Convergence

The Sanity Check: A series cannot converge unless |an| approaches zero for large n. The sanity check 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 finite

sum of defined values is always convergent. Trap the value of the remainder sum RN of a positive- term series between two integrals with N dependent limits and show that both integrals approach zero as N becomes large. Thus the remainder term goes to zero. This procedure is a special case of the comparison test.

∞ The Comparison Test for Convergence: Given that the series to test is ∑ak , find an absolutely k=1

∞ convergent series ∑bk such that |bk| > |ak| for all k greater than some N. If such a series can be k =1

∞ ∞ found, the series ∑ ak is absolutely convergent, and the series ∑ak converges to the same well- k =1 k=1 defined value for all re-orderings. The greater than some N appears because convergence is not determined by any finite number of terms

in the series; it is determined by the tail-end behavior RN.

∞ The Comparison Test for Divergence: Given that the series to test is ∑ak , find a positive term k=1

∞ divergent series ∑bk such that bk < |ak| for all k greater than some N. If such a series can be found, k =1

∞ then the series ∑ ak diverges. k =1

12/26/2007 Handout Series.Tank: Series, Sequences & Convergence SSC-10

Corollaries of the Comparison Test: ∞ a i.) If b is an absolutely convergent series, and n tends to a finite limit, ∑ k b k =1 n ∞ then the series ∑ak converges. k=1 ∞ ii.) If b is a divergent positive-term series, and an tends to a finite ∑ k bn k =1 ∞ limit (non-zero), then the series ∑ ak diverges. k =1

∞ The Ratio Test: Given that the series to test is ∑ak , form the sequence of the ratios of the next k=1 term to the current term. ⎧ <⇒1 absolutely convergent a ⎪ Limit n+1 →=⎨ 1 the test fails! n→∞ an ⎪ ⎩ >⇒1 fails to converge As the test involves the absolute values, the convergent case would be absolutely convergent. Note that in the case that the limit is 1, the test fails. Some other test must be applied to determine the convergence or divergence. The ratio test can be motivated by comparison test with the geometric series in the cases a < 1 and > 1. In the case of the ratio approaching 1, the test fails. That is: no definitive prediction can be made without additional information. One must appeal to more specialized 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 series n→∞

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 are many specialized tests for convergence. The discussion of additional tests is to be appended to the Tools of the Trade section just before the Problems as time permits or reason dictates.

12/26/2007 Handout Series.Tank: Series, Sequences & Convergence SSC-11

Sample Calculations/Applications of Convergence Tests: ∞ 1 ∞ 1 1.) The series ∑ can be rewritten as ∑ . This later positive term series dominates the series m=0 21m + m=1 21m − ∞∞11 1 ∞ 1 ∑∑= term by term. As the final ‘p = 1’ series diverges, so must ∑ by the comparison mm==1122mm m=0 21m + test.

∞∞(1)− m 2.) The series ∑∑am = is an alternating term series. Clearly |am+1|< |am| for all m, and mm==0021m +

Limit am = 0 . The series can be summed using a variety of tricks. In the handout, the sum is m→∞

∞ m π (1)− found to be 4 = ∑ . m=0 21m +

∞∞(1)− m 3.) The series a = is an alternating term series. Clearly |a |< |a | for all m, and ∑∑m 2 m+1 m mm==00(2m + 1)

Limit am = 0 . The series converges to a value defined to be Catalan’s constant (0.915965594… ). m→∞

∞∞1 4.) The series am = is an positive term series. Clearly, Limit am = 0 . ∑∑ 2 m→∞ mm==00(2m + 1) ∞∞11∞ 1 =+1 and is a positive term series that is dominated term by term ∑∑22∑ 2 mm==01(2mm++ 1) (2 1) m=1 (2m + 1) ∞∞11 ∞ 1 by = 1 . The final series, , converges, so all the series converge. ∑∑24 2 ∑ 2 mm==11(2mm ) ( ) m=1 ()m

∞ ∞ 1 2 1 2 = π ; = π ∑ 2 8 ∑ 2 6 m=1 (2m − 1) m=1 ()m

5.) Discuss the convergence of the binomial expansion for the inverse of (1+x).

∞∞∞ −1 nn(−− 1)( n 2)...( n −+ s 1) ( −−−−−− 1)( 1 1)( 1 2)...(s ) ()1+=x ∑∑xxs = ss=∑(1)− xs ss==00ss!!s=0 setn≡− 1

It is an alternating term series for positive x. As long as x > 0 the terms alternate in sign. For |x| < 1, |as+1|< |as|

12/26/2007 Handout Series.Tank: Series, Sequences & Convergence SSC-12 for all s, and Limit as = 0 . The series converges by the alternating terms series test for 0 < x < 1. Taking a s→∞ 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 so it must converge for all z such that |z| < R and diverge for all |z| > R. The alternating term series test indicated that R must be 1, and the geometric series result confirmed it.

∞∞ Exercise: Replace x by the complex value z. ∑∑(1)−−s xss→ (z ). ss==00

∞ Let z = a i where a is real and 0 < a < 1. What is ∑()−ai s ? Take the limit a → 1. Compare the result with s=0

∞ -1 s ½ (1 – i) = (1 + i) . Note than the series ∑ ()−i does not converge because |an| does not approach zero for s=0 large n. Mathematica 6: Sum[(- I)s ,{s,0,Infinity}]= 1/2-/2 ∞ − s s Mathematica 5.2: Sum − I , s, 0, Infinity Sum::div: Sum does not converge.= s=0

‚ H L 6.) Discuss the convergence@H ofL the 8binomial expansion

∞∞∞ −1 nn(−− 1)( n 2)...( n −+ s 1) ( −−−− 1)( 1 1)...(s ) ()1(−xx =∑∑−=)ss(−=x)∑xs ss==00ss!!s=0 setn≡− 1

The sanity check is valid: Limit as = 0 . The term by term ratio is x so, by the ratio test, the series converges s→∞ absolutely as long as |x| < 1. Could the ratio test have been applied in the previous example? Note that the ratio test result is valid in the complex plane so the series for (1-z)-1, converges everywhere inside a disk of radius one centered on z = 0.

z 7.) Discuss the convergence of the Taylor’s series expansion of e .

∞ zs zz23 zs ezz ==++++++∑ 1 ...... s=0 s! 22(3)s !

z th The (s + 1) to s term-by-term ratio is /(s + 1). The n term in the sum has index s = n – 1. The ratio test is:

12/26/2007 Handout Series.Tank: Series, Sequences & Convergence SSC-13 ⎧ <⇒1 convergent aa⎪ z Limit nn++11→=⎨ 1 test fails! Here, Limit= Limit →0for |z |finite nn→∞ →∞ s→∞ aann⎪ s ⎩ >⇒1 divergent n→∞ so, by the ratio test, the series converges absolutely for any finite |z|. A function with an unbounded or infinite radius of convergence is well-defined and analytic over the entire complex plane and is called an entire function.

8.) Discuss the convergence of the Taylor’s series expansion of [1 + 3x ] -1.

∞ s [1+=−=−+−+ 3xxxxx ]−12∑() 3 1 3 9 273 .... s=0

n 1 ()−3x + The absolute value of the term-by-term ratio is n =−33x = x . The ratio test is: ()−3x

⎧ <⇒1 convergent a ⎪ Limitn+1 ==→= Limit3|3| x x ⎨ 1test fails ! so, by the ratio test, the series converges nn→∞ →∞ an ⎪ ⎩ >⇒1 divergent 1 absolutely for |x|< /3.

Tools of the Trade:

Converting Sums to Integrals It is said that an integral is a sum of a huge number of small contributions, but some precision is required before the statement becomes useful. Beginning with a function f(t) and a sequence of

iN= t> values for t = {t1,t2,t3, ….,tN}, the sum f ()t does not represent the integral f ()tdt even if a ∑ i ∫t i=1 < great many closely spaced values of t are used. Nothing has been included in the sum to represent

iN= dt. One requires f ()tΔt where Δ=tt1 −t is the average interval between ∑ ii ii( 2 ) [ +11i− ] i=1

iN= sequential values of t values at ti. For well-behaved cases, the expression ∑ f ()ttiΔ i approaches the i=1 definition of an integral as the t-axis is chopped up more and more finely. As

12/26/2007 Handout Series.Tank: Series, Sequences & Convergence SSC-14 iN= illustrated below, in the limit Δt goes to zero, the sum ∑ f ()ttiΔ i approaches the area under the i=1

t> curve between t< and t>. That is: it represents f ()tdt provided the sequence of sums converges, ∫ t < and life is good. The theory of integration is not the topic of this passage. The goal is simply to remind you that the Δt must be factored out of each term that is being summed in order to identify the integrand.

f(t) f(tN)

f(ti) f(t1) f(tk)

Δt Δt

t

t1 t2 ti tk tN area = f(tk) Δt t< t>

In the Fourier series handout discussion of the inner product, the function h(t) = g(t) f(t) was

mN= –T +T considered at N equally spaced points between /2 and /2. This leads to the sum ∑ ht()m where m=1

the N points t=−TT − +m T for (1 ≤ m ≤ N) have equal spacing Δ=t T and are m ()22()N( N) N centered in each Δt wide interval. As the number of terms gets large, the sum must be divided by N

mN= to ensure a result finite. This leads to ht()1 . The rule for converting sums to integral ∑ m (N ) m=1

requires that Δ=t T be explicitly factored from each term in the sum. Thus N

mN==mN mN= +T /2 11T 1 ⎡⎤1 ht()mmm→=[]ht() [ht()]Δ t which becomes f ()tdt ∑∑()NTNT() () ∑() ∫−T /2 ⎣⎦()T mm==11 m=1 as N gets large and Δt small.

m ∞ ()−1 π Converting alternating term sums: ∑ = = 0.7853982 m=0 21m + 4

12/26/2007 Handout Series.Tank: Series, Sequences & Convergence SSC-15 m 7 ()−1 Anchor the evaluation. Sum through m = 7: ∑ = 0.7542680 m=0 21m +

m ∞ ()−1 The goal is to first estimate and then to provide bounds for the value of ∑ by m=8 21m + converting 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 term to 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 the terms. A trick is needed. Combine adjacent positive and negative terms in pairs. It is a trick; think about the sums below to verify that they are equal.

m ∞∞()−1 ⎡ 11⎤⎡∞ 2 ⎤ =−= ∑∑⎢ ⎥⎢ ∑2 ⎥ m==842m ++++ 1⎣ 4 1 4 3⎦⎣=4 16 16+ 3 ⎦ The last sum has the advantage that it is a sum of positive terms that vanish more rapidly than do the alternating terms in the original series. It has better convergence properties. To convert to an integral, Δ must be factored from each term, but takes on every integer greater than 4, the starting point, so Δ= 1. ∞∞⎡⎤⎡⎤2221∞ ⎡⎤19 ≡Δ≈=≈d ln 0.278064 ∑∑⎢⎥⎢⎥222∫ ⎢⎥ ==44⎣⎦⎣⎦16++ 16 3 16 ++ 16 34 16 ++ 16 3 4⎣⎦ 17

m 7 ()−1 ∞ 2 +≈d 0.7821 ∑ ∫ 2 m=0 2m +++ 14 16 16 3 π The exact value is /4 ≈ 0.785398. With patience one can do better.

∞∞211⎡⎤43N + Exercise: Show that: dd=− =u−1 du . Use the result to ∫∫2 ⎢⎥ ∫(4) NN16++ 16 3⎣⎦ 4 + 1 4 + 3 41 N+ evaluate the integral above.

Bounding the sum: For suitable choices of limits, the integrals can provide both upper and lower

12/26/2007 Handout Series.Tank: Series, Sequences & Convergence SSC-16 bounds for the sum. A careful sketch shows that for a monotone decreasing function F(n),

∞∞∞ Fxdx()<< Fn () Fxdx () . ∫∫NN∑ −1 nN=

The area of the shaded region is F(N), and F(x) as drawn is a monotone decreasing function. It follows that:

NN+1 F()xdxFN<< ( ) Fxdx () ∫∫NN−1 Exercise: Argue that, for a monotone decreasing function,

∞∞∞ Fxdx()<< Fn () Fxdx () ∫∫NN∑ −1 nN=

The shaded area is F(N) times 1. Make a sketch to NN+1 show that F()xdxFN<< ( ) Fxdx () given ∫∫NN−1 that F(x) is monotone decreasing.

mmm 7 ()−−11∞ 22∞ () 7 ()−1∞ +<<+dd ∑∑∫∫2 ∑2 mm==002mm++++ 14 16 16 3 2 1m=0 2m +++ 13 16 16 3

m ∞ ()−1 π or equivalently 0.7821 < ∑ < 0.7900. These results are consistent with a sum of /4 ≈ m=0 21m + 0.7854. A current estimate might be chosen to be:

m 7 ()−1 ∞ 21⎛⎞ ⎛17 ⎞ +≈+≈d 0.7542680 ln 0.7855587 ∑ ∫ 2 ⎜⎟ ⎜ ⎟ m=0 2m +++ 13.5 16 16 3 ⎝⎠4 ⎝ 15 ⎠ It seems that splitting the difference is reasonable so long as one remembers the bounds on the proven range.

Expansion/Identity Summary Double : n!! = n(n-2)(n-4) … {terminating with 2 or 1.}

Function Series Summation Rconv

12/26/2007 Handout Series.Tank: Series, Sequences & Convergence SSC-17 2 ∞ s df ⎛⎞1 d f 2 ⎛⎞1 df s f(x) fx()=+ fx ( ) 1 ()xx−− +()xx +... x − x - 0 () xx0000⎜⎟ 2 ∑⎜⎟ s x0 ()0 dx ⎝⎠2! dx s=0 ⎝⎠sdx! n 0 n-1 1 n(n-1) n-2 2 n n x y + n x y + /2! x y n n! ns− s (x + y) n(n-1) (n-2) n-3 3 ()x +=yx∑ ()!!nss− y - + /3! x y + ….. s=0 357 ∞ 21n+ xxx n x sin(x) x −+−+… ∑()−1 ∞ 3! 5! 7! n=0 ()21n + !

246 ∞ 2n xxx n x cos(x) 1−+−+… ∑()−1 ∞ 2! 4! 6! n=0 ()2!n xx3521762 x 7 x 7 tan(x) x ++ + + +… ½ π 3 15 315 2835 23456 ∞ n xxxxx n+1 x ln[1+x] x −+−+−+… ∑()−1 1 23456 n=1 n ∞ r rr(1)−−−23 rr (1)(2) r rr(−− 1)( r 2)...( r −+ s 1) s (1+x) 1.++rx x +x +. 1+ ∑ s! x 1 2! 3! s=1 n (1+x) If n is a positive integer, the series terminates. finite sum ⇒ no convergence issue xxx234 ∞ xn ex 1++x + + +… ∑ ∞ 23!4! n=0 n! ∞ ⎛⎞22mm+1 2345 m xx ix xxxx ∑()−+1 ⎜⎟i e 1+−ix − i + + i −… m=0 ()2!mn() 2+ 1! ∞ 23!4!4! ⎝⎠

xx3513ii 135 iii x 7 xmx32∞ (2− 1)!! m+1 Arcsin(x) x ++ − +… x ++∑ 1 62452467ii iii 6m=2 (2mm )!!(2+ 1) 357 ∞ 21n+ xxx n x Arctan(x) x −+−+… ∑()−1 1 357 n=0 ()21n + n π 11 1 π ∞ ()−1 Arctan(1/ ) −+ − +… + x 35 ∑ 21n+ 1 235xx x 2+n=0 ()21nx 2n xxx246 ∞ x cosh(x) 1++++… ∑ ∞ 2! 4! 6! n=0 ()2!n 21n xxx357 ∞ x + sinh(x) x ++++… ∑ ∞ 3! 5! 7! n=0 ()21n + ! xx3521762 x 7 x 7 tanh(x) x −+ − + −… 3 15 315 2835

Taylor’s Series expansion of f(x) about the argument value xo:

12/26/2007 Handout Series.Tank: Series, Sequences & Convergence SSC-18 2 ∞ s df ⎛⎞11d f 2 ⎛⎞d f s f ()xfx=+ ( ) 1 ()xx−− + ()xx +=... xx− 0 () xx0000⎜⎟ 2 ∑⎜⎟ s x0 ()0 dx ⎝⎠2! dx s=0 ⎝⎠s! dx Binomial Theorem: The theorem is an algebraic identity used to compute positive integer powers of a binomial.

n n ns− s n nnnnns! (− 1)(−−+ 2)...( 1) ()xy+=∑ ββns,, x y; ns ==()s = s=0 ()nss− !! s! TO APPLY, convert to the form (1 + x)n where |x| < 1 and use the form:

n nn(−−−−−−+ 1)23 nn ( 1)( n 2)nn ( 1)( n 2)...( n s 1) s (1++++xnxx )= 1 2! 3! x ++... s! x

Use the convention that only positive square roots considered for forms like az22+ as the

represent distances and that zzrzrz22=−=−; ( ) ;... unless you have a reason to do otherwise.

Euler's Identity: a relation between the exponential and the sinusoidal functions

also known as: The Euler-Lagrange identity eiiθ =+cosθ sinθ

Problems 1.) Express the following series in summation form. Interpret the … as it just keeps on going. 1111 11 1 1 1 a.) 1−+−+ −... b.) − +−+ −... 3579 4 9 16 25 36 xxx246 xxx357 c.) 1−+ −+… d.) x − +−+… 22346!⋅⋅ 23⋅⋅⋅⋅ 2345 7!

2.) It is assumed that the index increments by one from term to term in each series. (Parts a.) and b.) demonstrate a technique to effectively increment by 2 or 4.) Show the explicit form of the first five terms of the following series:

m ∞ ()−1 ∞ ⎡ 11⎤ a.) ∑ b.) ∑ ⎢ − ⎥ m=8 21m + =4 ⎣4143+ + ⎦ ∞ ⎡2 ⎤ ∞ (1)− n c.) d.) xn ∑ ⎢2 ⎥ ∑ =4 ⎣⎦16++ 16 3 n=0 n!

12/26/2007 Handout Series.Tank: Series, Sequences & Convergence SSC-19 ∞ (1)− n+1 ∞ (1)n − e.) f.) ∑ 2 ∑ 3 n=1 n n=1 n

3.) Find the radius of convergence for the expansion:

∞ n nn(−− 1)( n 2)...( n −[s −1] ) ()11+=+x ∑ xs s=1 s!

nnnnnn(1)(1)(2)−−− nn(−− 1)( n 2)...( n −−[ s 1] ) =+1 x12 +xx + 3+... + xs 1! 2! 3! s!

4.) The series expansion in x for ln(1+x) about x = 0 is:

∞ s ln 1+=x − 1 s+1 x ()()∑ ( s ) s=1 Find the circle of convergence for this expansion. What happens to ln( 1+ x) for x = -1? Comment on the circle of convergence in light of this observation.

5.) The series expansion in x for (1+x) -1 about x = 0 is:

∞ −1 ()1+=x ∑(1)− s xs s=0 Find the radius of the circle of convergence for this expansion. What happens to (1+x) -1 for x = -1?

∞ The radius of convergence for this problem is identical to that for ()1−=x −1 ∑ xs . Why? s=0 Comment on the circle of convergence in light of this observation. −1 2 3 4 5 6 7 8 Series 1+x , x, 0, 7 = 1− x + x − x + x − x + x −x + O x

∞ ∞ 1 2 1 2 6.) Use either of the sums H L = π and = π as a basis to establish the other. ∑ 2 8 ∑ 2 6 AH Lm=1 (2m 8− 1)

zs 7.) For the series below, show Limit as = 0 given that as = . Given a value of z, what is the s→∞ s!

12/26/2007 Handout Series.Tank: Series, Sequences & Convergence SSC-20 minimum value of s (smin) that ensures that |as+1| < |as| for s > smin ? This problem is the basis for the sanity check in the case of the series for

∞ znszz2 z ezz ==++++++∑ 1 ...... s=0 s! 22(3)s ! Find the radius of the circle of convergence of the series for ez.

8.) The solution method for: Legendre's Equation:

dP2 dP (1−−++xx2 ) 2 ( 1) P= 0 dx2 dx Leads to the series solutions:

⎧ ⎡⎤(1)+ 24[6(1)(1)−+][ +] ⎪ax0 ⎢⎥1− −+ x... ⎪ ⎣⎦(2)(1) 4! Peven(x) = ⎨ ∞ ⎪ 2i ax ⎪ ∑ 2i ⎩ i=0

⎧ ⎡⎤2(1)−+ 35[12)−+ ( 1)][ ( + 1)] ⎪ax1 ⎢⎥+− x x+... ⎪ ⎣⎦(3)(2) 5! Podd(x) = ⎨ ∞ ⎪ 21i+ ax ⎪ ∑ 21i+ ⎩ i=0 nn(1)(1+ −+ ) Use the recursion relation aa= and study the convergence of these series n+2 n (2)(1)nn++ as a function of x using the ratio test. The ratio test fails for x = ± 1. Other tests establish that the general series diverges for x = ± 1. Finite results can be salvaged at x = ± 1 provided is restricted a particular of values. Given those values, the infinite series terminate after a finite number of terms leading to a set of polynomials. Give the value spectrum for that leads to series truncated as finite sums.

∞ 1 2 10 1 9.) It has been claimed that = π . Show that ≈ 1.54977. Use the integral method to ∑ 2 6 ∑ 2 m=1 ()m m=1 ()m ∞ 1 2 bound the value of R = and thereby demonstrate that π is a possible limit. Continue the 10 ∑ 2 6 m=11 ()m 100 1 procedure to better bound the sum. Start with ≈1.6349839002 and establish bounds for ∑ 2 m=1 ()m 2 R . π ≈ 1.6449340. 100 6

12/26/2007 Handout Series.Tank: Series, Sequences & Convergence SSC-21

10.) A ball is dropped from rest at a height H and falls with a uniform acceleration g downward traveling a distance H to hit the floor the first time. (Ignore air resistance.) What is the time required for the ball to fall the distance H? At that point the ball strikes a floor and rebounds upward with a coefficient of restitution e. The speed of the ball just after the bounce is e times its speed just before the bounce. What is the maximum height that the ball reaches on the first bounce? How long does it take to reach that height? Find the total distance traveled and the total time elapsed during the first bounce. Assuming the same coefficient of restitution for every bounce, construct sums to represent the total distance that the ball travels and the total time elapsed since the ball was released. Sum the series for an infinite number of bounces.

⎡⎤12++eH2 ⎡1e⎤ ANSWER: DH==; T total ⎢⎥2 total ⎢ ⎥ ⎣⎦11−−eg⎣ e⎦

d d d n d n 11.) Compute dx [cosh(x )] and dx [sinh(x )]. Develop expressions for dxn [cosh(x )] and dxn [sinh(x )] for all positive integer values: n = 0, 1, 2, 3, … . Use these results, and generate the Taylor’s series expansions about xo = 0 for hyperbolic cosine and hyperbolic sine. Find the radii of convergence for hyperbolic cosine and hyperbolic sine.

∞∞ 12.) Parameter Calculus Used to Sum Series: Show that the series nann= ad a . Use this ∑ da ∑ nn==00

∞ ∞ to find ∑ nan . Extend the procedure to compute ∑nn(1)− an . n=0 n=0

13.) Convert a number in repetend form (eg. x = 2.191919... , a repeating decimal) into a ratio of two integers. The decimal expression for x repeats every 2 digits. Form (102 x – x). The result is an (102 x – x) integer value. Hence x = / (102 – 1). is x expressed as a ratio of integers. The problem may be

∞ m 1 attacked using series techniques as well. 2.191919… = 2 + 19 ∑()100 . Use the geometric series m=1

12/26/2007 Handout Series.Tank: Series, Sequences & Convergence SSC-22 ∞∞ mma 1 198 1 217 aaa=== . So 2.1919…. = /99 + 19 ( /99) = /99. Employ this geometric series ∑∑1−aa−1 −1 mm==10 approach to express 2.7128128128………… as a rational number, the ratio of integers.

14.) After making careful measurements of the electrostatic potential at point on the z axis for large

∞ q m z, a student concluded that the potential could be expressed as: a . Sum the series to ∑ ( z m+1 ) 4πε 0 m=0 show that this could be the potential of a single charge and identify its location.

References: 1. The Wolfram web site: mathworld.wolfram.com/ 2. K. F. Riley, M. P. Hobson and S. J. Bence, Mathematical Methods for Physics and Engineering, 2nd Ed., Cambridge, Cambridge UK (2002). 3. Donald A. McQuarrie, Mathematical Methods for Scientists and Engineers, University Science Books, Sausalito, CA (2003).

12/26/2007 Handout Series.Tank: Series, Sequences & Convergence SSC-23