Math 9B Fial Study Aid Jue 6, 20 Geeral advise. Get plety of practice. There s a lot of material i this sectio - try to do as may examples ad as much of the homework as possible to get some practice. Just readig this review or just readig the chapter i the text/otes wo t prepare you - you have to actually do some examples to make it sik i. Chapter.7 has a buch of good examples for practice o applyig the covergece/divergece tests. Older material. The fial is comprehesive. Make sure you leave some time to review the earlier chapters - the past exams ad your ow midterms are useful study aids. Keep i mid the examples you spet time o your homework doig which have t appeared o a midterm yet. These are likely topics for the fial. Try to review some of the past homeworks too. Differetiatio recap. There s a fair amout of differetiatio required i this chapter - you ll eed it to apply L Hopital s theorem ad also to compute Taylor series (from scratch). Brush up o your skills - i particular make sure you kow the quotiet rule, chai rule ad product rule - they re all i the text if you wat to look them up. Be careful applyig the covergece tests. I the exam a lot of people will apply the covergece tests from this chapter icorrectly. Make sure you kow uder what coditios a give test ca be used ad what coclusios it geerates. A lot of the tests are icoclusive i certai situatios. If you re ot sure if you re applyig the test correctly read the statemet of the theorem agai or go ad see a TA / Ed to check. I the exam be sure to state which theorem / test you are usig ad why it applies (i.e. show that the coditios of the test are met). Chapter. - Sequeces. Defiitios Sequece. A sequece is a assigmet of a real umber a to each atural umber N (N stads for the atural umbers or the coutig umbers,2,3,4,...). The otatios we use to deote a sequece are {a, a 2, a 3,... }, {a }, {a } N or (most commoly) just {a }. Limits, covergece ad divergece. A sequece {a } has a it L, which we deote a L as or a = L, if we ca make the terms a as close to L as we like by makig sufficietly large. A sequece may or may ot have a it. If it does the we say that the sequece coverges or the it exists. If it does t we say the sequece diverges or the it does ot exist. Factorial,!, 0! If is a atural umber, the symbol! (proouced factorial ) is short had for! = ( ) ( 2) ( 3) 2.
Theory For example, 5! = 5 4 3 2. By covetio (ad slightly couter-ituitively), we defie 0! =. ( ) Note that! oly makes sese if is a atural umber or zero:! is osese. 2 The associated fuctio theorem. If f(x) = L ad f() = a for every N the x a = L. The squeeze theorem. If a b c for all N ad a = c = L the b = L. Cotiuous fuctio theorem. If a = L ad the fuctio f is cotiuous at L the Remarks. f(a ) = f(l). Fidig its of sequeces. Throughout this chapter you ll eed to kow how to compute its. You ca state, without proof, the followig results i the exam: ( ) = 0 (l ) ( a ) = 0 (a ) ( e ) = 0 (e ) = x/ = (x > 0) x! = 0 x = 0 ( x < ) for ay costat a >. I additio to these mai results there are two approaches that are ofte used to compute its. (i) L Hopital s theorem. L Hopital s theorem states that if f(x) is differetiable at c (usually c will be i our examples) ad f(x) = g(x) = 0, or f(x) = g(x) = ±, the x c x c x c x c f(x) x c g(x) = f (x) x c g (x) provided this it exists or is ifiite. Iformally, we use L Hopital s rule for its of the form 0/0 or ± / ±. We may also apply L Hopital s theorem to tackle its of the form 0, ad 0 by usig a bit of algebraic maipulatio to reduce it to a it of the form 0/0 or ± / ±. Here s a example: x si(/x) it is of the form 0 x = x si(/x) /x x 2 cos(/x) = x x ( ) 2 = x cos ( = cos x x ( x )) = cos(0) =. it is of the form 0/0 applyig L Hopital s Page 2
Note that by the associated fuctio theorem, the above calculatio shows that the sequece {l /} N (which techically is t a differetiable fuctio) coverges to the same it 0. (ii) Limits of quotiets. The other commo case you ll eed to kow is how to take its such as 3 33 + 2 5. 2 + 8 To approach these, first determie the degree of the umerator ad the deomiator. If the umerator has larger degree tha the deomiator the the it teds to ifiity ad does ot exist. Coversely, if the deomiator has larger degree tha the umerator the the it will be zero. I the case where both umerator ad deomiator have the same degree, pick the highest power of from the umerator ad the deomiator (takig ito accout radicals) ad divide the umerator ad deomiator by this. I the example above, both the umerator ad deomiator have the same degree () so: 3 33 + 2 5 2 + 8 = ( 3 33 + 2 5 ) / (2 + 8)/ = 3 3 + 5 3 2 + 8 = 3 3 2. Cotiuous fuctio theorem. Almost all the fuctios we see i this course are cotiuous. This theorem says it s ok to push its through cotiuous fuctios. For example, sice l is cotiuous we have ( ) ( + l = l + ) = l () = 0. Chapter.2 - Series. Defiitios. Ifiite series ad covergece. Cosider a sequece {a }. Addig together all the a : gives what we call a series. You may thik of it as a ifiite sum. If we just add together the first terms i the sequece we obtai the th partial sum, deoted s : If the it s exists the s = a a i = a + a 2 + + a. i= s = s is called the sum ad the series a is said to coverge. We use the short-had to express this fact. If the it s does ot exist the the series is said to diverge. s = a Page 3
Theory. Geometric series. The specific series ar = a + ar + ar 2 + is called the geometric series. It coverges (i.e. you get a umerical aswer for this ifiite sum ) if r < ad diverges if r. If it coverges, we ca calculate its sum usig the formula: The divergece test. For ay series ar = a r. a, if a 0 the the series diverges. Sums of series. If a ad b are coverget series the so too are c a (for ay costat c), (a + b ) ad (a b ) with: c a = c a (a + b ) = a + b (a b ) = a b. The harmoic series. This will be metioed agai later (it is a p-series with p = ). The series diverges. = + 2 + 3 + 4 + Telescopig series... Yeah, kow how to do these! I ll fill out some commets here whe I have time... Remarks. Sequeces vs. series. Be sure to differetiate betwee the two - a series is a ifiite sum, a sequece is just a ifiite list of (ordered) terms. Whe to use the divergece test. (Always!) A good proportio of this chapter (ad hece your exam!) is dedicated to the methods we use to decide whe a series will or will ot coverge. This test oly addresses the later case but is by far the easiest to implemet of all the tests you ll come across. Therefore, the divergece test should always be the first thig you check whe you re attemptig to test for (covergece)/divergece. Note that if a = 0 the the divergece test is icoclusive - i particular, it does ot tell you that the series will coverge (there are lots of examples, such as the harmoic series, where a = 0 but the series still diverges). Page 4
The geometric sum is v.useful! Fidig the sum of a series is usually extremely difficult, eve i the cases where we kow that it coverges (ad therefore the sum must exist). (I geeral it is a impossible task, the best we ca usually achieve is a approximatio usig umerical methods - see later). The geometric series however, is oe of the few istaces whe the sum ca be calculated exactly. If a questio asks you to compute the sum of a ifiite series (exactly, rather tha by approximatios) you will either have to use this formula or the series will be a telescopig example. Notice that the geometric series is characterized by the fact that each term a i the series is obtaied by multiplyig the previous term a by some value r. Wheever a series has this property it is a geometric series ad you ca maipulate it so that the formula above may be used to fid its sum (provided r < ). As a fial ote, you should be aware that the geometric series cotais its ow cogergece / divergece test - calculatig the ratio of successive terms i a geometric series will give you the value r. A geometric series always coverges if r < ad always diverges if r (- compare this to the ratio test below). Recogize sums of series. Throughout this chapter it will sometimes be useful to split series ito a sum of two series or to factor out a costat to facilitate determiig covergece ad/or to use the geometric series formula above. There is a example cotaied i the compariso test remarks that uses this property. Note also that just as for differetiatio ad itegratio, multiplicatio by a costat has little effect o a series - you ca just factor it out to the frot ad forget about it. Chapter.3 - The itegral test. Theory. Itegral test. This gives a practical method for determiig if a series coverges or diverges. If f is a cotiuous, positive, decreasig fuctio o [, ) such that f() = a, the: (i) if (ii) if f(x) dx is coverget the f(x) dx is diverget the The p-series. The series a is coverget, a is diverget. is very importat. So importat it has its ow ame: the p-series. You should memorize the fact that it coverges for p > ad diverges if p. Note that for p = it is just the harmoic series, which as we already oted, diverges. You ca prove the coditios uder which this series coverges ad diverges usig the itegral test above - it boils dow to the fact that the improper itegral p coverges if p > ad diverges if p. x p dx Remarks. Whe to use the itegral test. Use it whe a) you are asked to determie the covergece or divergece of a fuctio, b) whe the divergece test (Chap.2) has bee icoclusive ad c) whe the a terms look like a fuctio you ca itegrate. Page 5
Fidig f ad start values. The fuctio f(x) you choose for the itegral test will be obvious - just replace s with x s i the a terms. Also, it does ot matter if the series fails to start at =, simply alter the itegral accordigly. Both of these remarks are illustrated i determiig the covergece / divergece of e 2. Method: we just check whether the itegral =3 coverges or diverges. 3 xe x2 dx It s oly a test! The itegral test just checks for divergece or covergece of a series. If the series covergeces the the sum is ot the value of the improper itegral. If you are specifically asked to calculate the sum of a series you should probably be usig the geometric series formula or applyig the telescopig method. Chapter.4 - Compariso tests. Theory. The compariso test. If a ad b are series with positive terms the: (i) if b coverges ad a b for all, the a coverges, (ii) if a diverges ad a b for all, the a diverges. Note that this is exactly the same setup for the compariso test for improper itegrals. The it compariso test. If a ad b are series with positive terms ad Remarks. a = c b where 0 < c <, the either both series diverge or both series coverge. Whe to use them. First off, check that the divergece test (chap.2) does t apply ad that you ca t itegrate the terms i the series (otherwise just use the itegral test). The compariso tests are probably the hardest tests to implemet i practice - oly apply them whe you have to. The idea behid these compariso tests is to compare the series we are give to a simpler series which we kow either coverges or diverges. The oly series you are expected to kow the covergece / divergece of are the p-series ad the geometric series, so these are the oly two series you will use i the compariso. Of the two, the p-series compariso is more commoly used - or it is i your homework ad examples at least... We therefore use the compariso test whe the series approximately looks like (!) either a geometric series or (more commoly) a p-series. To be more specific, ad to give a commo example, you would attempt to make use of a p-series compariso whe determiig the covergece / divergece of series of the form: f() g() where the quotiet is positive for all values of ad f() ad g() are either a) polyomials i, b) polyomials raised to some (costat) power (possibly /2 - i.e. a square root) or c) sequeces that Page 6
are bouded for 0 (for example si(), cos 2 (), e etc...). You would cosider usig a geometric series compariso if the terms i the series you re iterested i cotai powers of (although keep i mid that the root test or ratio test may be more appropriate i this istace - see later). For example: 2 3 +, + + 2 3 + 2 +, 6 ( + ) e, ( ) si 2 () 2 should all be compared to p-series (ote that i the last two examples e ad si 2 () are bouded above by ). These series: 4 + 2 3, + 3 + 4, + si 0 should all be compared to geometric series. Which series to compare to? If you are makig a p-series compariso, the choose the highest power i both the umerator ad the deomiator of the fuctio. The simplify ad try to use this as your p-series compariso (it wo t always work but it gives you a startig poit). For example, for + + 2 3 + 2 + 8 2 for the compariso (which coverges). I other cases it is ot so clear cut. use 2 3 = 2 8 2 2 4 = 2 2 Compariso vs it compariso... Which test to use? As a geeral rule of thumb, the it test is usually easier to implemet ad saves you havig to fudge costats to get the iequalities to work out. It may however leave you with a difficult it to calculate (review the remarks i Chap. if ecessary). If it is obvious that the terms of oe series are larger or smaller tha the other, the compariso test will probably be quicker ad easier. Whe i doubt pick oe, see if it works out ad if ot try the other. Practice a few examples so that you re used to usig both approaches. If either approach works see the commets below! Chapter.5 - Alteratig series test. Defiitios Alteratig series. A alteratig series is a series i which cosecutive terms alterate betwee positive ad egative: Theory. ( ) b = b b 2 + b 3 b 4 + b 5... with b > 0. i= Alteratig series test. If a alteratig series ( ) + b satisfies (i) b + b for all, (ii) b = 0, the the series is coverget. Page 7
Remarks. Whe to use the alteratig series test. This test should be cosidered for determiig the covergece of ay series that is alteratig. However, see the commets below cocerig absolute covergece... I practice you should check the secod coditio of the alteratig series test first. If coditio two does ot hold (i.e. b 0) the you do t have to go ay further - the series will diverge by the divergece test (Chap.2). O the other had, if the first coditio does ot hold this does ot show that the series diverges - it just meas the test is icoclusive. The reaso for this is that the first coditio ca be weakeed slightly - see below. Weakeed alteratig series test. This is ot too importat but you should be aware that the first coditio of the alteratig series test ca be weakeed slightly - all that is importat is that b + b evetually - i.e. for all sufficietly large or, i other words, for all bigger tha some large umber N. Showig b + < b. Differetiatio ca be used to show that b + < b. For example, cosider the series ( ) + 2 3 +. To show that the sequece b = 2 /( 3 + ) is decreasig, it suffices to show that the related fuctio f(x) = x 2 /(x 3 + ) is decreasig (for sufficietly large x values). The fuctio f(x) is decreasig wherever f (x) < 0. Sice f (x) = x ( 2 x 3) (x 3 + ) 2 we see that f(x) is decreasig for 2 x 3 < 0 or, i other words, for x > 3 2. We coclude that b is decreasig (i.e. b + b ) for all 2. Therefore (by the commets above) the alteratig series test applies. Chapter.6 - Absolute covergece, the ratio ad root tests. Defiitios. Absolutely coverget. A series a is called absolutely coverget if a is coverget. Coditioally coverget. A series a is called coditioally coverget if it is coverget but ot absolutely coverget. I other words, a is coditioally coverget if a is coverget but a is diverget. Theory. Absolute covergece implies covergece. If a series is absolutely coverget the it is coverget. The ratio test. Let a be ay series. (i) If a + a = L <, the the series a is absolutely coverget. (ii) If a + a = L > or a + a =, the the series a is diverget. (iii) If a + a =, the the ratio test is icoclusive. Page 8
The root test. Let a be ay series. Remarks. (i) If a = L <, the the series a is absolutely coverget. (ii) If a = L > or a =, the the series (iii) If a =, the root test is icoclusive. a is diverget. Absolute covergece stroger tha covergece. Be aware that if a fuctio is ot absolutely coverget the it may still be coditioally coverget. Alteratives to the alteratig series test. Because absolute covergece implies (usual) covergece the just because a series is alteratig does t ecessarily mea we have to use the alteratig series test. We might istead fid it easier to apply the ratio test or the compariso test or the itegral test to a i order to show absolute covergece - which i tur implies covergece. To prove that a alteratig series is coditioally coverget you eed to show that a) a is diverget (so that the series does ot coverge absolutely), ad b)that a is coverget. The latter will require the alteratig series test. Whe to use the ratio test. Series that ivolve factorials or costats beig raised to a power of - for example: 2 k k!, ( ) 2 4, ( 0),! are all coveietly tested usig the ratio test. Whe ot to use the ratio test. The ratio test will ot work o ratioal fuctios or algebraic ratioal fuctios (ratioal fuctios with square roots etc). For these problems use the compariso test for a suitable p-series - see whe to use the compariso tests. Whe to use the root test. Pretty much oly whe a is of the form (b ) (ad is t a geometric series!), for example: ( + ) ( 2),, ( ) 5 2. + Chapter.7 - Stratergy for testig series. This sectio is very useful - it s like chapter 7.5 for itegratio; there are lots of problems ad you have to work out which divergece / covergece test is appropriate for the series. Read through the stratergy for testig series o page 72. This is excellet practice for the exam sice whe you come to take your test it is ulikely that you ll be told which test to use. Try ad have a look at as may of these problems as possible. They test you o all the covergece / divergece tests you have leart so far ad get you used to quickly spottig which test is appropriate. Practice is the key! Chapter.8 - Power series. Defiitios. Power series. A power series about a is a series of the form c (x a) = c 0 + c (x a) + c 2 (x a) 2 + c 3 (x a) 3 +... =3 =2 Page 9
where x is a variable, c are costats - (called the coefficiets of the power series) ad a is a costat. Note that a could be zero i which case we have a power series about the origi : c (x 0) = c x = c 0 + c x + c 2 x 2 + c 3 x 3 +... Radius of covergece, iterval of covergece. Give a power series Theory. c (x a), the iterval of covergece is the set of all x values for which the series coverges. By the theorem below, this will always be a iterval; hece the ame iterval of covergece. The radius of covergece is the largest umber R such that the series coverges for every x o the iterval (a R, a + R) (i.e. for every x with a R < x < a + R). Covergece of power series. For a power series about a, i.e. Remarks. c (x a), there are oly three possibilities for the values of x that will make the power series coverge: (i) The series oly coverges whe x = a. I this case the iterval of covergece is [a, a] (i.e. just the poit a) ad the radius of covergece is R = 0. (ii) The series coverges for every value of x. I this case the iterval of covergece is (, ) (i.e. all real umbers) ad we say the radius of covergece is ifiite : R =. (iii) The series coverges for all x less tha a distace r from a - that is, for all x such that x a < r - ad diverges for all x a distace greater tha r from a - that is for all x such that x a > r. I this case the radius of covergece is R = r ad the iterval of covergece is (a r, a + r) or (a r, a + r] or [a r, a + r) or [a r, a + r] depedig o whether or ot the power series coverges at x = a r ad/or x = a + r (which we call the edpoits of the radius of covergece or just the edpoits ). Commets o the defiitios. What is a power series ad how is it differet to a usual series? The key differece is that a power series is a fuctio of x. The way this fuctio works is that for each value of x you iput, it outputs a series (i the usual sese): x value power series series For example, cosiderig the power series (x 3), Page 0
the pluggig i say, x = 4, gives the series ( ). It is importat to make this distictio betwee power series ad series. The iterval of covergece is exactly the domai of this fuctio - the values of x that output a coverget series. Determiig the radius of covergece. To fid the iterval / radius of covergece of a power series c (x a) just apply the ratio test (or occasioally you ll eed the geometric series test) to the terms a = c (x a) i the power series. For example, for the power series the a = x /! ad so a + a = x + ( + )!! x = x + = 0 for every x. The ratio test therefore implies the power series coverges for every x (recall that the ratio test says that the series will coverge if a + /a < - we ve show this happes for all values of x). This meas the radius of covergece is R = ad the iterval of covergece is (, ). Take care with edpoits! Whe usig the geometric series test the ed poits of the iterval of covergece are ot icluded. Whe usig the ratio test you must test the ed poits of the iterval of covergece maually - the power series may coverge there or it may ot. To determie this, plug the edpoits ito your power series ad, usig a appropriate covergece / divergece test from chapters.2 to.6, determie whether the resultig series (it s just a ormal series oce you have chose a x value) coverges or diverges. Here s a example. Fid the iterval of covergece of the power series We set a = (x + 2) /3 + ad fid that a + a = x! (x + 2) 3 +. ( + ) x + 2 = 3 Therefore, by the ratio test, the power series will coverge whe x + 2 3 < x + 2. 3 - that is, whe x + 2 < 3 (ote that this says that the radius of covergece is 3). Aother way of expressig this iterval is 5 < x <. The problem is, the ratio test is icoclusive at the edpoits x = 5 ad x = ad so we must check these maually. This meas checkig the two series ( 5 + 2) 3 + = ( 3) 3 + ad ( + 2) 3 + = (3) 3 + for covergece or divergece idepedetly. The best test to use o both of these is the divergece test - for both series we have a 0 ad so either series coverges. Hece the iterval of covergece is ( 5, ) (i iterval otatio). Before we tested the ed poits, ay of the followig itervals could have bee the iterval of covergece: ( 5, ), [ 5, ), ( 5, ], [ 5, ] (recall that square brackets idicate that you iclude the edpoit - i.e. ( 5, ] is the iterval 5 < x ). Page
Chapter.9 - Represetatios of fuctios as power series. Theory. Differetiatig ad itegratig power series. If a power series c (x a) has radius of covergece R > 0 the the fuctio defied by f(x) = c (x a) = c 0 + c (x a) + c 2 (x a) 2 + c 3 (x a) 3 +... is differetiable ad itegrable o the iterval of covergece (or domai of f(x)) (a R, a + R). Furthermore, (i) f (x) = (ii) c (x a) = c + 2c 2 (x a) + 3c 3 (x a) 2 +... f(x) dx = c (x a) + (x a) 2 = C + c 0 (x a) + c + 2 + c 2 (x a) 3 3 +.... ad the radius of covergece of (i) ad (ii) is still R - although covergece at the edpoits may chage (the iterval of covergece could be differet)! This theorem says that o the iterval of covergece you fid the derivative (respectively itegral) of the power series by just differetiatig (resp. itegratig) term by term. Remarks. I ve combied the remarks of this sectio with those from the ext sectio - see below. Chapter.0 - Taylor (ad Maclauri) series. Theory. Taylor series theorem. Suppose f(x) is ay fuctio that ca be differetiated ifiitely may times a poit x = a. The for some radius of covergece R, f(x) has a power series expasio about a: f(x) = c (x a) x a < R Remarks where the coefficiets c are give by the formula c = f () (a).! The power series above is called the Taylor series expasio of f(x) at a. If we take the Taylor series expasio at a = 0 we get what is called the Maclaure series of f(x): f(x) = f () (0)! x = f(0) + f (0)! x + f (0) x 2 +... x < R. 2! Notatio. The otatio f () (a) i the Taylor series expasio meas the th derivative of f evaluated at x = a: f () (a) = d f dx (x) = d f x=a dx (a). Page 2
Why Taylor series? Taylor series are importat because the theory of itegratio is icomplete: ulike differetiatio, there are cotiuous (perfectly ice lookig fuctios!) that either we ca t itegrate or simply do t have a atiderivate. Oe example is e x2. Taylor series, which ca be itegrated o their iterval of covergece, (see the differetiatig ad itegratig power series theorem from the previous chapter) provide oe approach to partially get aroud this problem. Some examples. Computig Taylor series is ot hard, it s all just computatio (although you may wat to revise your differetiatio skills!). Here are some commo Taylor series (cetered at 0) alog with their radius of covergece - you should be able to deduce all these straight from the defiitio: x = x = + x + x 2 + x 3 +... R = e x = = + x! + x2 2! + x3 3! +... R = si x = ( ) x2+ (2 + )! = x x3 3! + x5 5! x7 7! +... R = cos x = ( ) x2 (2)! = x2 2! + x4 4! x6 6! +... R =. You probably wo t be expected to memorize these but you should be able to derive them. As a example of calculatig Taylor series expasios, cosider the problem of fidig the Maclaure series expasio (i.e the Taylor series expasio about x = 0) of f(x) = l x +. We have: Hece the Taylor series of this fuctio is f(x) = l x + f (x) = x + f (x) = (x + ) 2 f (x) = 2 (x + ) 3 f (4) (x) = 3 2 (x + ) 4... f () (x) = ( )+! (x + ). f () (0)! x = f(0) + = ( ) + x ( ) +!! (0 + ) x = x x 2 + x 3 x 4 +.... This series is geometric with commo ratio r = ( x). The Taylor series therefore coverges if, ad oly if, r = x <. Short-cuts for computig Taylor expasios. The method above for computig Taylor series is very time-cosumig. I your exam you may be give a Taylor series ad the asked to use it to Page 3
compute the expasio of a related fuctio. For example, usig the cos x expasio above, we have ( ) ( ( x cos 2 x2 = x ( ) 2 x2) ) ( 2 ) = x ( ) x 4 (2)! (2)! 2 2 = ( ) x4+ (2)! 2 2. Sice the radius of covergece of the Taylor series of cos x is R = the the radius of covergece of this power series is also R =. However, this is a special case - i geeral the ew series will ot have the same radius of covergece as the origial series. Suppose, for example, the radius of covergece of cos x had bee R = 8 i the example above (it s ot, but just suppose), the the series coverges for x < 8. We replaced this x with a 2 x2 ad so the ew series coverges whe 2 x2 < 8, that is, for x < 4. Hece the radius of covergece of the ew series we derived would be R = 4. Sorry that s a bad example, it s gettig late! Haha... More short-cuts! Aother commo way of geeratig Taylor series (without doig all the log calculatios) is to use the differetiatio ad itegratio of power series theorem from the last chapter. Note that the theorem tells us that radius of covergece remais uchaged after itegratig or differetiatig (although the iterval of covergece may be differet - i.e. the ed poits may o loger be coverget / diverget). Here s a example: The Taylor series of /( x) about 0 (give above) has radius of covergece R =. Thus for x < we have x = + x + x2 + x 3 + x 4 +.... (ote that this is just the usual formula for the sum of the geometric series). Differetiatig this gives the Taylor expasio of /( x) 2 o x < : Similarly, itegratig gives ( x) 2 = + 2x + 3x2 + 4x 3 + 5x 4 +... l x = C + x + x2 2 + x3 3 + x4 4 +... which is valid for x <. The radius of covergece of both these last two series is equal to the first, R =. Calculatig ifiite sums. If we have a Taylor series expasio c x for a fuctio f(x) we ca use it to fid the value of specific ifiite series. For ay value x = k such that k belogs to the iterval of covergece the c k = f(k). For example, the power series e x = coverges for all x. Therefore, choosig x = we have x!! = +! + 2! + 3! + = e. Page 4