INFINITE SEQUENCES AND SERIES

INFINITE SEQUENCES AND SERIES

INFINITE SEQUENCES AND SERIES I geeral, it is difficult to fid the exact sum of a series. We were able to accomplish this for geometric series ad the series /[(+)]. This is because, i each of these cases, we ca fid a simple formula for the th partial sum s. Nevertheless, usually, it is t easy to compute lim s

INFINITE SEQUENCES AND SERIES So, i the ext few sectios, we develop several tests that help us determie whether a series is coverget or diverget without explicitly fidig its sum. I some cases, however, our methods will eable us to fid good estimates of the sum.

INFINITE SEQUENCES AND SERIES Our first test ivolves improper itegrals.

INFINITE SEQUENCES AND SERIES.3 The Itegral Test ad Estimates of Sums I this sectio, we will lear how to: Fid the covergece or divergece of a series ad estimate its sum.

INTEGRAL TEST We begi by ivestigatig the series whose terms are the reciprocals of the squares of the positive itegers: = = + + + + + 2 2 2 2 2 2 2 3 4 5... There s o simple formula for the sum s of the first terms.

INTEGRAL TEST However, the computer-geerated values give here suggest that the partial sums are approachig ear.64 as. So, it looks as if the series is coverget. We ca cofirm this impressio with a geometric argumet.

INTEGRAL TEST This figure shows the curve y = /x 2 ad rectagles that lie below the curve.

INTEGRAL TEST The base of each rectagle is a iterval of legth. The height is equal to the value of the fuctio y = /x 2 at the right edpoit of the iterval.

INTEGRAL TEST Thus, the sum of the areas of the rectagles is: + + + + +... = 2 3 4 5 2 2 2 2 2 2 =

INTEGRAL TEST If we exclude the first rectagle, the total area of the remaiig rectagles is smaller tha the area uder the curve y = /x 2 for x, which is the value of the itegral (/ x 2 ) dx I Sectio 7.8, we discovered that this improper itegral is coverget ad has value.

INTEGRAL TEST So, the image shows that all the partial sums are less tha Therefore, the partial sums are bouded. + dx = 2 x 2 2

INTEGRAL TEST We also kow that the partial sums are icreasig (as all the terms are positive). Thus, the partial sums coverge (by the Mootoic Sequece Theorem). So, the series is coverget.

INTEGRAL TEST The sum of the series (the limit of the partial sums) is also less tha 2: = = + + + +... < 2 2 2 2 2 3 2 4 2

INTEGRAL TEST The exact sum of this series was foud by the mathematicia Leohard Euler (707 783) to be π 2 /6. However, the proof of this fact is quite difficult. See Problem 6 i the Problems Plus, followig Chapter 5.

INTEGRAL TEST Now, let s look at this series: = = + + + + + 2 3 4 5...

INTEGRAL TEST The table of values of s suggests that the partial sums are t approachig a fiite umber. So, we suspect that the give series may be diverget. Agai, we use a picture for cofirmatio.

INTEGRAL TEST The figure shows the curve y = /. However, this time, we use rectagles whose tops lie above the curve. x

INTEGRAL TEST The base of each rectagle is a iterval of legth. The height is equal to the value of the fuctio y = / x at the left edpoit of the iterval.

INTEGRAL TEST So, the sum of the areas of all the rectagles is: + + + + +... = 2 3 4 5 =

INTEGRAL TEST This total area is greater tha the area uder the curve y = / the itegral x for x, which is equal to (/ x) dx However, we kow from Sectio 7.8 that this improper itegral is diverget. I other words, the area uder the curve is ifiite.

INTEGRAL TEST Thus, the sum of the series must be ifiite. That is, the series is diverget.

INTEGRAL TEST The same sort of geometric reasoig that we used for these two series ca be used to prove the followig test. The proof is give at the ed of the sectio.

THE INTEGRAL TEST Suppose f is a cotiuous, positive, decreasig fuctio o [, ) ad let a = f(). The, the series is coverget if ad oly if the improper itegral is coverget. = a f ( x) dx

THE INTEGRAL TEST I other words, i. If is coverget, the f ( x) dx is coverget. = a ii. If f ( x) dx is diverget. is diverget, the = a

NOTE Whe we use the Itegral Test, it is ot ecessary to start the series or the itegral at =. For istace, i testig the series we use 4 2 ( x 3) dx = ( 3) 4 2

NOTE Also, it is ot ecessary that f be always decreasig. What is importat is that f be ultimately decreasig, that is, decreasig for x larger tha some umber N. The, a = N is coverget. a = So, is coverget by Note 4 of Sectio.2

INTEGRAL TEST Example Test the series or divergece. 2 = + for covergece The fuctio f(x) = /(x 2 + ) is cotiuous, positive, ad decreasig o [, ).

INTEGRAL TEST Example So, we use the Itegral Test: x t dx = lim dx + x + 2 t 2 = lim ta t lim ta x ] = t t π π π = = 2 4 4 t π 4

INTEGRAL TEST Example So, 2 /( x + ) dx is a coverget itegral. Thus, by the Itegral Test, the series /( 2 + ) is coverget.

INTEGRAL TEST Example 2 For what values of p is the series = p coverget?

INTEGRAL TEST Example 2 If p < 0, the lim(/ p ) = If p = 0, the lim(/ p ) = I either case, p lim(/ ) 0 So, the give series diverges by the Test for Divergece (Sectio.2).

INTEGRAL TEST Example 2 If p > 0, the the fuctio f(x) = /x p is clearly cotiuous, positive, ad decreasig o [, ).

INTEGRAL TEST Example 2 I Sectio 7.8 (Defiitio 2), we foud that : x p dx Coverges if p > Diverges if p

INTEGRAL TEST Example 2 It follows from the Itegral Test that the series / p coverges if p > ad diverges if 0 < p. For p =, this series is the harmoic series discussed i Example 7 i Sectio.2

INTEGRAL TEST To use the Itegral Test, we eed to be able to evaluate Therefore, we have to be able to fid a atiderivative of f. f ( x) dx Frequetly, this is difficult or impossible. So, we eed other tests for covergece too.

p-series The series i Example 2 is called the p-series. It is importat i the rest of this chapter. So, we summarize the results of Example 2 for future referece as follows.

p-series Result The p-series = is coverget if p > ad diverget if p p

p-series The series = Example 3 a = + + + +... 3 3 2 3 3 3 4 3 is coverget because it is a p-series with p = 3 >

p-series Example 3 b The series = = + + + +... 2 3 4 /3 3 3 3 3 = = is diverget because it is a p-series with p = ⅓ <.

NOTE We should ot ifer from the Itegral Test that the sum of the series is equal to the value of the itegral. I fact, = 2 = π 6 2 whereas x 2 dx = Thus, i geeral, = a f ( x) dx

INTEGRAL TEST Example 4 l = Determie whether the series coverges or diverges. The fuctio f(x) = (l x)/x is positive ad cotiuous for x > because the logarithm fuctio is cotiuous. However, it is ot obvious that f is decreasig.

INTEGRAL TEST Example 4 So, we compute its derivative: f '( x) (/ xx ) l x l x = = x 2 x 2 Thus, f (x) < 0 whe l x >, that is, x > e. It follows that f is decreasig whe x > e.

INTEGRAL TEST Example 4 So, we ca apply the Itegral Test: 2 l x t l x (l x) dx = lim dx = lim x x t 2 t 2 (l t) = lim = t 2 t Sice this improper itegral is diverget, the series Σ (l )/ is also diverget by the Itegral Test.

ESTIMATING THE SUM OF A SERIES Suppose we have bee able to use the Itegral Test to show that a series a is coverget. Now, we wat to fid a approximatio to the sum s of the series.s

ESTIMATING THE SUM OF A SERIES Of course, ay partial sum s is a approximatio to s because lim = s How good is such a approximatio?

ESTIMATING THE SUM OF A SERIES To fid out, we eed to estimate the size of the remaider R = s s = a + + a +2 + a +3 + The remaider R is the error made whe s, the sum of the first terms, is used as a approximatio to the total sum.

ESTIMATING THE SUM OF A SERIES We use the same otatio ad ideas as i the Itegral Test, assumig that f is decreasig o [, ).

ESTIMATING THE SUM OF A SERIES Comparig the areas of the rectagles with the area uder y = f(x) for x > i the figure, we see that: R = a+ + a+ 2 +... f ( x) dx

ESTIMATING THE SUM OF A SERIES Similarly, from this figure, we see that: R... ( ) = a+ + a+ 2 + f x dx +

ESTIMATING THE SUM OF A SERIES Thus, we have proved the followig error estimate.

REMAINDER ESTIMATE (INT. TEST) Estimate 2 Suppose f(k) = a k, where f is a cotiuous, positive, decreasig fuctio for x ad a is coverget. If R = s s, the + f ( x) dx R f ( x) dx

REMAINDER ESTIMATE Example 5 a. Approximate the sum of the series / 3 by usig the sum of the first 0 terms. Estimate the error ivolved. b. How may terms are required to esure the sum is accurate to withi 0.0005?

REMAINDER ESTIMATE Example 5 I both parts, we eed to kow f ( x) dx With f(x) = /x 3, which satisfies the coditios of the Itegral Test, we have: dx = lim x 2 x 3 t 2 = lim + = t 2 t 2 2 t 2 2 2

REMAINDER ESTIMATE Example 5 a = s 3 0 = + + + +.975 3 3 3 3 2 3 0 As per the remaider estimate 2, we have: R dx x 2(0) 200 = = 0 0 3 2 So, the size of the error is at most 0.005

REMAINDER ESTIMATE Example 5 b Accuracy to withi 0.0005 meas that we have to fid a value of such that R 0.0005 Sice R dx = x 2 3 2 we wat 2 2 < 0.0005

REMAINDER ESTIMATE Solvig this iequality, we get: Example 5 b 2 > = 000 or > 000 3.6 0.00 We eed 32 terms to esure accuracy to withi 0.0005

REMAINDER ESTIMATE Estimate 3 If we add s to each side of the iequalities i Estimate 2, we get s f ( x) dx s s f ( x) dx + + + because s + R = s

REMAINDER ESTIMATE The iequalities i Estimate 3 give a lower boud ad a upper boud for s. They provide a more accurate approximatio to the sum of the series tha the partial sum s does.

REMAINDER ESTIMATE Example 6 Use Estimate 3 with = 0 to estimate the sum of the series = 3

REMAINDER ESTIMATE Example 6 The iequalities i Estimate 3 become: s + dx s s + dx 0 3 0 x 0 x 3

REMAINDER ESTIMATE From Example 5, we kow that: Example 6 Thus, dx = x 2 3 2 s + s s + 2() 2(0) 0 2 0 2

REMAINDER ESTIMATE Example 6 Usig s 0.97532, we get:.20664 s.202532 If we approximate s by the midpoit of this iterval, the the error is at most half the legth of the iterval. Thus, = 3.202 with error < 0.0005

REMAINDER ESTIMATE If we compare Example 6 with Example 5, we see that the improved estimate 3 ca be much better tha the estimate s s. To make the error smaller tha 0.0005, we had to use 32 terms i Example 5, but oly 0 terms i Example 6.

PROOF OF THE INTEGRAL TEST We have already see the basic idea behid the proof of the Itegral Test for the series / 2 ad /.

PROOF OF THE INTEGRAL TEST For the geeral series a, cosider these figures.

PROOF OF THE INTEGRAL TEST The area of the first shaded rectagle i this figure is the value of f at the right edpoit of [, 2], that is, f(2) = a 2.

PROOF OF THE INTEGRAL TEST Estimate 4 So, comparig the areas of the rectagles with the area uder y = f(x) from to, we see that: a + a + + a f ( x) dx 2 3 Notice that this iequality depeds o the fact that f is decreasig.

PROOF OF THE INTEGRAL TEST Likewise, the figure shows that: Estimate 5 f ( x) dx a + a + + a 2

PROOF OF THE INTEGRAL TEST Case i If is coverget, the Estimate 4 gives f ( x) dx sice f(x) 0. i i= 2 a f ( x) dx f ( x) dx

PROOF OF THE INTEGRAL TEST Therefore, Case i s = a + a a + f ( x) dx i i= 2 = M, say Sice s M for all, the sequece {s } is bouded above.

PROOF OF THE INTEGRAL TEST Case i Also, s = s + a s + + sice a + = f( + ) 0. Thus, {s } is a icreasig bouded sequece.

PROOF OF THE INTEGRAL TEST Case i Thus, it is coverget by the Mootoic Sequece Theorem (Sectio.). This meas that a is coverget.

PROOF OF THE INTEGRAL TEST If f ( x) dx is diverget, Case ii the f ( x ) dx as because f(x) 0.

PROOF OF THE INTEGRAL TEST However, Estimate 5 gives: Case ii = i i= f ( x) dx a s Hece, s. This implies that s, ad so a diverges.