# 6.3 Testing Series With Positive Terms

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1 6.3. TESTING SERIES WITH POSITIVE TERMS Testig Series With Positive Terms 6.3. Review of what is kow up to ow I theory, testig a series a i for covergece amouts to fidig the i= sequece of partial sums S = kow that lim S = a i ad testig it for covergece. We i= a i. So, if the sequece of partial sums coverges, i= so will the series ad they coverge to the same umber. If the sequece of partial sums diverges, so will the series. I theory, this souds like the problem is solved, sice we kow how to study the covergece of sequeces. I practice though it is ot as easy. I geeral, we will ot have a ice formula for S, therefore it may be diffi cult to determie if it coverges. This techique does work i some examples. Whe the sequece of partial sums caot be used, there are several alteratives which we list here: If the sequece is a kow series, use your kowledge of that series. This is why, as you lear ew series, it is importat to remember them. So far, we kow the followig series: Harmoic series: Geometric series: diverges otherwise.. It diverges. i ar i. It coverges to i= i= a r if r < ad If the sequece is ot kow, the first test that should be applied is the test for divergece. It says that if lim 0, the the series a i will diverge. If this limit is zero, the test provides o i= coclusio. Other tests must be used. Remark 6.3. It should be obvious to the reader that covergece of a series is ot affected by the first fiitely may terms of the series. I other words, if coverges, so will a i, i= a i,..., i=3 i= a i where p is ay positive iteger. The i=p same is true for divergece. What will chage however, is the sum. Obviously, if a i

2 308 CHAPTER 6. INFINITE SEQUENCES AND SERIES a i coverges say to L,that is if a i = L, the we also kow that a i L. i= It is easy to fid what it is though. i= a i = L a. (why?) i= The rest of this documet presets some of the tests which ca be used to test series for covergece. We first begi with tests which oly apply to series havig positive terms. Later o we will look at tests for series havig mixed terms. The rest of this sectio is oly cocered with series of positive terms that is series of the form a i where a i 0 for every i. i= 6.3. The Itegral Test This test compares a series to a itegral (improper). It establishes that uder certai coditios, covergece of a series ca be established by studyig the covergece of a improper itegral. It eve goes further. The value of the itegral ca also be used to approximate the ifiite series. However, ulike for geometric series, this test will ot give us a exact value for coverget ifiite series. Theorem 6.3. (Itegral test) If f is a cotiuous, positive ad decreasig fuctio o [, ) ad a = f () the a i coverges if ad oly if f (x) dx i= coverges. Proof. Usig the otatio S = a i with a i = f (i) where f satisfies the i= coditios of the itegral test, it is easy to see that + f (x) dx S a + f (x) dx Sice a i 0, {S } is a icreasig sequece. If the itegral coverges, lim f (x) dx is fiite hece a + f (x) dx is also fiite. Thus {S } is bouded above. Sice it is also icreasig, it must coverge. Hece a i coverges. O the other had, if the itegral diverges, + f (x) dx as. Hece lim S = thus a i diverges. i= Remark The coditios i the itegral test do ot have to hold startig at =, as log as they hold from some poit o. However, if they hold o a iterval of the form [ 0, ) for some positive iteger 0 the oe will study the covergece of the itegral 0 f (x) dx istead of f (x) dx. i= i=

3 6.3. TESTING SERIES WITH POSITIVE TERMS 309 Remark Before applyig this test, make sure that the series satisfies all the coditios of the theorem. Remark The theorem does ot idicate what the series coverges to. You should ot assume that the value of the itegral ad the series are the same. Example Test = l for covergece. The fuctio defiig the geeral term of this series is f (x) = l x x. It is cotiuous ad positive o [, ). To determie if it is decreasig, we study the sig of its derivative. f (x) = l x. Sice the deomiator is a square, x it is always positive. So the sig of the derivative is uiquely determied by its umerator. f (x) < 0 l x < 0 < l x l x > x > e Therefore, f is ot decreasig o [, ) but it is o [3, ), which is good eough. Remember, this property oly eeds to hold from some poit o, which it does here. Havig verified that all the hypotheses of the theorem were satisfied, we ca ow use the itegral test. We study the covergece of a improper itegral. 3 l x x dx = lim t 3 = lim t = t l x x dx [ (l t) The itegral diverges, therefore the series diverges. ] (l 3) 3 l x x dx. This is Remark The above itegral was evaluated usig the substitutio u = l x therefore du = dx x. Example Test i= i p for covergece. If p < 0, the lim i i p =, so the series diverges. If p = 0, the i p = i 0 =.

4 30 CHAPTER 6. INFINITE SEQUENCES AND SERIES hece, lim =. The series also diverges. If p > 0, the the fuctio i ip f (x) = x p is cotiuous, decreasig (sice its deomiator is icreasig) ad positive o [, ). So, istead of studyig the covergece of the series, we study the covergece of the itegral dx. From a theorem about improper xp itegrals, we kow that this itegral coverges if p > ad diverges otherwise that is i this case whe 0 < p. Combiig all the cases, we see that the series coverges whe p > ad diverges otherwise. Remark This proves that the harmoic series, a p-series with p =. i=, diverges sice it is i The above example is very importat. Its result should be remembered. Series which appear i the above example have a special ame. Defiitio A p-series is a series of the form Example 6.3. Here are some examples of p-series. i= i p.. 3. i= i= i= (p = ) i i (p = ) i 3 (p = 3 ) I example 6.3.8, we proved the followig: Theorem 6.3. A p-series coverges if p >. It diverges otherwise. Example Test i i= for covergece. This is a p-series with p =, therefore it coverges. Example Test i= i for covergece. This is a p-series with p =, therefore it diverges.

5 6.3. TESTING SERIES WITH POSITIVE TERMS Compariso Tests The idea here is to compare a ukow series to a kow series. Series such as the harmoic series, geometric series ad p-series are used a lot. The questio is, give a ukow series, how do we kow what to compare it to. Series which look like a p-series or a geometric series should be compared with such series. For example, = +, thus = + should be compared with = +. Similarly, = because for large, should be compared to. I additio, the followig facts are ofte used ad should be remembered. If a, b, c are positive umbers the b > c implies that a b < a c l x < x < e x Oe of two compariso tests ca be used: Theorem (Stadard compariso test) Suppose that a ad b are two series with positive terms such that a b for every from some poit o. The, If b coverges, so does a (i.e. if the series with larger terms coverges, so will the series with smaller terms). If a diverges, so does b (i.e. if the series with smaller terms diverges, so will the series with larger terms). Ituitively this theorem should make sese. Sice we have series of positive terms, either the series is fiite, i which case it coverges or it is ifiite, i which case it diverges. So, if 0 a b the 0 a b So, we see that if the series of larger terms, b coverges that is is fiite, the a must also be fiite hece coverge. Equivaletly, if a diverges, that is is ifiite, the b must also be ifiite that is diverge. These are the oly two situatios uder which we ca coclude. Remark It is easier to remember the theorem i this form: Give two series of positive terms. If the series of larger terms coverges, so does the series of smaller terms. Equivaletly, if the series of smaller terms diverges, so does the series of larger terms. The idea here is that if we suspect that the give series coverges, we eed to fid a series with larger terms which we kow coverges. Coversely, if we

6 3 CHAPTER 6. INFINITE SEQUENCES AND SERIES suspect the give series diverges, we eed to fid a series with smaller terms which we kow diverges. It is also importat to ote that the iequality does ot have to hold for every. It just has to hold from some poit o. Example Test for covergece. + 5 For large, + 5 behaves like. So, we compare + 5 to which is a p-series with p =, thus it coverges. At this poit, we caot coclude yet that the give series coverges. For this, we eed to prove that the terms of. This is easy to see because + 5 are smaller tha the terms of ad therefore follows that + 5 coverges. Example Test l. By the stadard compariso test, it for covergece. Sice l for 3, it follows that l By the stadard compariso test, l Example Test +. We kow that diverges. diverges. for covergece. For large, + behaves like. We also kow that coverges (geometric series with r = ). Thus we suspect our series coverges. Sice +, we ca use the stadard compariso test to coclude that it does. Example Test 5 for covergece. Sice 5 behaves like, we suspect our series coverges sice coverges. However, 5. So, we caot use the stadard compariso test. The ext compariso test addresses this issue. Theorem 6.3. (Limit compariso test) Suppose that a ad b are two series with positive terms such that L = lim. If 0 < L <, the two series behave alike. a b exists or is the:. If L = 0 the covergece of b implies covergece of a. Divergece of a implies divergece of b.

7 6.3. TESTING SERIES WITH POSITIVE TERMS If L = the covergece of a implies covergece of b. Divergece of b implies divergece of a. Ituitively, the theorem should make sese. If lim a b = 0 the this idicates that b is much larger tha a. Thus, b is the series with larger terms. We ca use remark to help us remember what the theorem allows us to a coclude. Similarly, if lim =, the a is much larger tha b. So, this b time a is the series with larger terms. Example 6.3. Test for covergece. 5 This is the series we could ot hadle with the stadard compariso test. Agai, we compare it to. This time, we look at lim 5 = lim = 5 also co- Sice, we coclude by the limit compariso test that 5 verges Usig the Itegral Test to Estimate Sums As poited out above, the itegral test ca be used to study the covergece of certai series. But, the itegral ad the series do ot have the same value. However, the itegral ca be used to fid the error whe approximatig the sum of a series by its sequece of partial sums. Let us itroduce some otatio. Let us assume that we have a series a i, which satisfies the coditios of the itegral test. Assume we have prove that this series coverges. We would like to kow what it coverges to. Call S = a i Obviously, we caot add up all the terms; there is a ifiite amout. i= But we kow from the test for divergece that because the series coverges, lim a = 0. Therefore, the terms we are addig become smaller ad smaller. Evetually, they will be so small that they will almost ot cout. This suggests that we could approximate this ifiite sum by addig a fiite umber of terms ad igore the remaiig oes. I other words, we could approximate S = a i by S = a i. Of course, the two will ot be exactly equal. Whe we i= i= approximate somethig with somethig else, it is importat to kow the error we are makig. Let us deote the error by R. I other words, R = S S. i=

8 34 CHAPTER 6. INFINITE SEQUENCES AND SERIES It turs out that ot oly we ca fid the error, we ca also fid how may terms we eed to add if we eed the error to be less tha a certai umber. This is the ext theorem, which we will state without proof. The proof ca be foud i ay calculus book. Theorem If a i coverges ad satisfies the coditios of the itegral i= test ad if we set R = S S, the + f (x) dx R f (x) dx (6.) ad therefore S + f (x) dx S S + f (x) dx (6.) + We illustrate with a few examples how this theorem ca be used to fid the error whe we approximate S with S as well as to fid out how may terms must be added so that the error is less tha a certai amout. The secod idea is extremely importat. We kow that whe we build somethig, whether it is a bridge, or tiy parts for a computer, we ca ever build these parts to the exact measuremets. Hece, there is the idea of "tolerace", that is the error we are allowed for the part to still fuctio. Obviously, the tolerace for a bridge is much larger tha for computer parts. This idea of tolerace exists i pretty much everythig we do. So, it may happe that a real problem has a ifiite series as its solutio. Because we may ot be able to fid exactly what the ifiite series is, we may have to approximate it. The tolerace of the problem we are solvig will tell us the maximum error our approximatio ca have. If you thik about it, this is pretty amazig. We do ot kow what the exact sum of the series is, yet we ca tell how far from the exact value our approximatio is. Example Cosider the series. Suppose we approximate this series with S 0 (the sum of the first 0 terms). What will be the error ad what i3 i= ca we say about the exact value of i 3? i= Let S deote the exact sum. We are approximatig S with S 0. Usig the otatio above, the error is R 0. From the theorem, we see that x 3 dx R 0 0 x 3 dx

9 6.3. TESTING SERIES WITH POSITIVE TERMS 35 We begi by computig the itegrals. Therefore, Ad t dx = lim x3 t x 3 dx [ ] t = lim t x [ = lim t t + ] = x 3 dx = () 0 = 4 = x 3 dx = (0) = 00 =.005 It follows that if we approximate S by S 0, the error R 0 will satisfy R I other words, the error is at least but o greater tha.005. I fact, we ca eve do better, usig the secod iequality of the theorem, ad 0 the fact that S 0 = i 3 = = , we see that i= S 0 + x 3 dx S S x 3 dx S S Thus, if we approximate S by the midpoit of these two values, we have S

10 36 CHAPTER 6. INFINITE SEQUENCES AND SERIES Ad the error is less tha half the differece betwee these two values that is R Remark What the above example suggests is that we ca approximate S by usig the midpoit of the iterval S + + f (x) dx, S + f (x) dx. I doig so, the error is less tha half the width of the same iterval, that is R f (x) dx + f (x) dx This provides a better approximatio as the example below will cofirm. (6.3) Example Approximate the series with a error less tha.00. i i= I other words, we wat to fid (how may term we should add) so that R <.00.. Usig the first approach. We wat to approximate i i= with S, so that R.00. From the theorem ad equatio 6., we kow that R dx. Thus, if we solve x begi by computig the itegral. dx.00, what we wat will follow. We x t dx = lim x t x dx [ ] t = lim t x [ = lim + ] t t To have withi.00 we use =.00, we must have.00 = 000. To approximate i 000 i= i. Ideed, i= i = 6 π = ad i= 000 i= i =

11 6.3. TESTING SERIES WITH POSITIVE TERMS With this approach, we had to add the first 000 terms of the series.. Usig equatio 6.3, we wat to fid so that This happes whe + ( + ) x dx + x dx.00. Usig your calculator, you ca try values of util the iequality is satisfied. Whe =, ( + ) = (4) () = >.00. But whe =, ( + ) = (44) (3) = <.00. This meas that that if we approximate S by the midpoit of S + 3 f (x) dx, S + the error will be less tha.00. You may thik this is more work, but it is ot the case. Most of the work is i addig the terms of the series. Above, we added the first 000 terms. Here, we oly add the first terms. Furthermore, ad 3 S = i= i = f (x) dx = 3 f (x) dx = = = Therefore, the midpoit of the iterval S + 3 f (x) dx, S + f (x) dx f (x) dx,

12 38 CHAPTER 6. INFINITE SEQUENCES AND SERIES is S + f (x) dx + S + f (x) dx 3 = = This is withi the desired accuracy Thigs to Kow Studets should kow ad be able to use the followig tests Itegral test p-series, what they are, whe they coverge Stadard ad limit compariso tests Studets should also be able to approximate the sum of a series, ad determie the error of the approximatio Problems. Suppose that Σa ad Σb are series with positive terms ad Σb is kow to be coverget. (a) If a > b for all, what ca be said about Σa? Why? (b) If a < b for all, what ca be said about Σa? Why?. It is importat to distiguish betwee b ad = b. What ame is give to the first series? To the secod? For what value of b doe the first series coverge? For what value of b doe the secod series coverge? 3. Use the itegral test to determie if the itegrals below coverge or diverge. = (a) (b) = = 4 e 4. Use a compariso test to determie if the series below coverge or diverge. (a) = + +

13 6.3. TESTING SERIES WITH POSITIVE TERMS 39 (b) (c) (d) = = = cos cos 5. Determie whether the series below coverge or diverge usig ay of the tests studied so far. (a) (b) e (c) (d) (e) (f) (g) (h) (i) = = = = l ( ) si + ( ) = = = = + si Fid the values of p for which 7. Cosider the series = 4. = (l ) p is coverget. (a) Use the sum of the first 0 terms to estimate estimate? =.How good is this 4

14 30 CHAPTER 6. INFINITE SEQUENCES AND SERIES (b) How may terms should we add if we wat the error to be less tha 0 6? 8. The meaig of the decimal represetatio of a umber 0.d d d 3 d 4... (where d i is a iteger betwee 0 ad 9) is that 0.d d d 3 d 4... = d 0 + d 0 + d d Show that this series always coverge Aswers. Suppose that Σa ad Σb are series with positive terms ad Σb is kow to be coverget. (a) If a > b for all, what ca be said about Σa? Why? othig. (b) If a < b for all, what ca be said about Σa? Why? Σa coverges by the stadard compariso test.. It is importat to distiguish betwee b ad = b. What ame is give to the first series? To the secod? For what value of b doe the first series coverge? For what value of b doe the secod series coverge? b : p series, coverges for b <. = = b : geometric series, coverges whe b <. = 3. Use the itegral test to determie if the itegrals below coverge or diverge. (a) (b) 4 = coverges. = e coverges. 4. Use a compariso test to determie if the series below coverge or diverge. (a) + + = coverges.

15 6.3. TESTING SERIES WITH POSITIVE TERMS 3 (b) (c) (d) cos 3 = coverges. 3 = coverges. + cos = coverges. 5. Determie whether the series below coverge or diverge usig ay of the tests studied so far. (a) coverges. (b) e = coverges. (c) l = diverges. 5 (d) + 3 = coverges. + (e) = diverges. + (f) 4 + = coverges. ( ) (g) si = diverges. + ( ) (h) = Coverges. + si (i) 5. = coverges.

16 3 CHAPTER 6. INFINITE SEQUENCES AND SERIES 6. Fid the values of p for which coverges if ad oly if p >. 7. Cosider the series = 4. = (l ) p is coverget. (a) Use the sum of the first 0 terms to estimate estimate? R =.How good is this 4 (b) How may terms should we add if we wat the error to be less tha 0 6? 70 terms. 8. The meaig of the decimal represetatio of a umber 0.d d d 3 d 4... (where d i is a iteger betwee 0 ad 9) is that 0.d d d 3 d 4... = d 0 + d 0 + d d Show that this series always coverge. hit: compare it to a geometric series.

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