10.5 Positive Term Series: Comparison Tests Contemporary Calculus 1

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1 0. Positive Term Series: Compariso Tests Cotemporary Calculus 0. POSITIVE TERM SERIES: COMPARISON TESTS This sectio discusses how to determie whether some series coverge or diverge by comparig them with other series which we already ow coverge or diverge. I the basic Compariso Test we compare the two series term by term. I the more powerful Limit Compariso Test, we compare its of ratios of the terms of the two series. Fially, we focus o the parts of the terms of a series that determie whether the series coverges or diverges. Compariso Test Iformally, if the idividual terms of our series are smaller tha the terms of ow coverget series, the our series coverges. If our series is larger, term by term, tha ow diverget series the our series diverges. If the idividual terms of our series are larger tha the terms of a coverget series or smaller tha the terms of a diverget series, the our series may coverge or diverge the Compariso Test does ot tell us. Compariso Test Suppose we wat to determie whether a coverges or diverges. (a) If there is a coverget series c with 0 < c for all, the a coverges. (b) If there is a diverget series d with d > 0 for all, the a diverges. Proof: Sice all of the terms of the, c, ad d series are positive, their sequeces of partial sums are all mootoic icreasig. The proof compares the partial sums of the various series. (a) Suppose that 0 < c for all ad that c coverges. Sice c coverges, the the partial sums t c approach a fiite it: t L. For each, s t (why?) so s t L, ad the sequece { s } is

2 0. Positive Term Series: Compariso Tests Cotemporary Calculus 2 bouded by L. Fially, by the Mootoe Covergece Theorem, we ca coclude that { s } coverges ad that a coverges. (b) Suppose that d > 0 for all ad that d diverges. Sice d diverges, the the partial sums u d approach ifiity: u. The s u so the sequece of partial sums { s } diverges ad the series a diverges. The Compariso Test requires that we select ad compare our series agaist a series whose covergece or divergece is ow, ad that choice requires that we ow a collectio of series that coverge ad some that diverge. Typically, we pic a p series or a geometric series to compare with our series, but this choice requires some experiece ad practice. Example : Use the Compariso Test to determie the covergece or divergece of (a) 2 ad (b) Solutio: For these two series it is useful to compare with p series for appropriate values of p. (a) For all, < 2, ad 2 coverges (P Test, p 2) so 2 coverges. + 3 (b) For all, >. Sice diverges (P Test, p ), we coclude that + 2 diverges. Practice : Use the Compariso Test to determie the covergece or divergece of (a) ad (b)

3 0. Positive Term Series: Compariso Tests Cotemporary Calculus 3 Example 2: A studet has show algebraically that 2 < 2 < for all 2. From this iformatio ad the Compariso Test, what ca the studet coclude about the covergece of the series $ 2 2? Solutio: Nothig. The Compariso Test oly gives a defiitive aswer if our series is smaller tha a coverget series or if our series is larger tha a diverget series. I this example, our series is larger tha a coverget series, 2, ad is smaller tha a diverget series,, so we ca ot coclude aythig about the covergece of $ 2 2. However, we ca show that if 2 the 2 < Sice 2 2 coverges (P Test), we ca coclude that $ 2 2 coverges. Next i this sectio we preset a variatio o the Compariso Test that allows us to quicly coclude that $ 2 2 coverges. Limit Compariso Test The exact value of the sum of a series depeds o every part of the terms of the series, but if we are oly asig about covergece or divergece, some parts of the terms ca be safely igored. For example, the three series with terms / 2, /( 2 + ), ad /( 2 ) coverge to differet values, 0.64, , + 2 $ , but they all do coverge. The + ad i the deomiators effect the value of the fial sum, but they do ot effect whether that sum is fiite or ifiite. Whe is a large umber, the values of /( 2 + ) ad /( 2 ) are both very close to the value of / 2, ad the covergece or divergece of the series 2 2 ad $ ca be predicted from the covergece or divergece of the series 2 2. The Limit Compariso Test states these ideas precisely.

4 0. Positive Term Series: Compariso Tests Cotemporary Calculus 4 Limit Compariso Test Suppose > 0 for all, ad we wat to determie whether a coverges or diverges. If there is a series b so that b L, a positive, fiite value, the a ad b both coverge or both diverge. Idea for a proof: The ey idea is that if very large, b L so L. b ad a L. N b L is a positive, fiite value, the, whe is b. If oe of these series coverges, the so does the other. If oe of these series diverges, the so does the other. Whe is a relatively small umber, the ad b values may ot have a ratio close to L, but the first few values of a series do ot effect the covergece or divergece of the series. A proof of the Limit Compariso Test is give i a Appedix after the Practice Aswers. N Example 3: Put 2, b 2 +, ad c 2 ad show that b ad that c. Sice 2 2 ad coverges (P Test, p 2) we ca coclude $ 2 2 both coverge too. Solutio: b / 2 /( 2 + ) so L is positive ad fiite. Similarly, c / 2 /( 2 ) so L is positive ad fiite.

5 0. Positive Term Series: Compariso Tests Cotemporary Calculus 2 Practice 2: (a) Fid a p series to it compare with put b L, a positive, fiite value. Does ad fid a value of p so that b p ad coverge? (b) Fid a p series to compare with 3 4. Does 3 4 coverge? The Limit Compariso Test is particularly useful because it allows us to igore some parts of the terms that cause algebraic difficulties but that have o effect o the covergece of the series. Usig Domiat Terms To use the Limit Compariso Test we eed to pic a ew series to compare with our give series. Oe effective way to pic the ew series is to form the ew series usig the largest power of the variable (domiat term) from the umerator ad the largest power of the variable (domiat term) from the domiat term i the umerator deomiator. The domiat term series cosists of domiat term i the deomiator. The the Limit Compariso Test allows us to coclude that the origial series coverges if ad oly if the domiat term series coverges. Example 4: For each of the give series, form a ew series cosistig of the domiat terms from the umerator ad the deomiator. Does the series of domiat terms coverge? 2 (a) (b) (c) Solutio: (a) The domiat terms of the umerator ad deomiator are 2 ad 2 4, respectively, so the 2 domiat term series is which coverges (P Test, p 2 ). (b) The domiat terms of the umerator ad deomiator are 4 ad 3/2, respectively, so the domiat term series is 4 3/2 4 /2 which diverges (P Test, p /2 ). (c) The domiat terms of the umerator ad deomiator are 23 ad 26, respectively, so the 23 domiat term series is 26 3 which coverges (P Test, p 3 ).

6 0. Positive Term Series: Compariso Tests Cotemporary Calculus 6 Usig the Limit Compariso Test to compare the give series with the domiat term series, we ca coclude that the give series (a) ad (c) coverge ad that the give series (b) diverges. Practice 3: For each of the give series, form a ew series cosistig of the domiat terms from the umerator ad the deomiator. Does the series of domiat terms coverge? Do the give series coverge? 4 (a) (b) 2 (c) Experieced users of series commoly use domiat terms to mae quic ad accurate judgmets about the covergece or divergece of a series. With practice, so ca you. PROBLEMS I problems 2 use the Compariso Test to determie whether the give series coverge or diverge. 2. cos ( ) si( ) cos( j ) j j 6. j arcta( j ) j 3/2 7. l( ) ( ) + 2.! ( ). I problems 3 2 use the Limit Compariso Test (or the N th Term Test) to determie whether the give series coverge or diverge j 7 j w + w ( ) ( + 2 ) 3 8. ( arcta( ) ) w ( + w ) w 2. j ( j ) j

7 0. Positive Term Series: Compariso Tests Cotemporary Calculus 7 I problems use domiat term series to determie whether the give series coverge or diverge j 4j + 3 j2j 4 + 7j ( + 3 ) ( arcta( 3 ) 2 ) ( ) j j 3 + 4j 2 j 2 + 3j Puttig it all together I problems 3 use ay of the methods from this or previous sectios to determie whether the give series coverge or diverge. Give reasos for your aswers j j + j3j 4 + 2j (3 + 2 ) ( arcta( 2 ) 3 ) ( ) j j 2 + 4j j si( ) si( π ) je j + j (2 + 3) + 9 ( + 3 ) 2 4. ( ta( 3 ) 2 + ) si 2 ( ) 47. si 3 ( ) 48. cos 2 ( ) 49. cos 3 ( j ) 0. ta 2 ( ). ( 2 )

8 0. Positive Term Series: Compariso Tests Cotemporary Calculus 8 Review for Positive Term Series: Coverge or Diverge State whether the give series coverge or diverge ad give reasos for your aswer. You may eed ay of the methods discussed so far as well as some igeuity. 3 R. 33 R2. + cos( j ) 2 j3 j 2 R3. w3 + si( w 3 ) R4. ( /3 ) R. e R6. si 2 ( w w ) (Hit: for 0 x, si(x)<x) R7. si( 3 ) (see the R6 hit) R8. j cos 2 ( 2 j ) R9. + cos( ) 3 R0.. (3 + l()) R. jj. (3 + l(j)) 2 R2. 4. arcta( ) R3. 4. arcta( ) R4. 3 l( ) 3 R. l() 2 j R6. ( 2j + 3 j ) j R R8. si( ) si( + ) R R20. R2. / Practice Aswers Practice : (a) For > 3, 0 < 2 < so 0 < 2 < ad 3 /2 diverges (P test, p /2) so 3 (b) For >, > 2 > 0 so < 2. 2 > /2. 2 diverges. 32 ( 2 3 ) which is a coverget geometric series ( r /2) so coverges.

9 0. Positive Term Series: Compariso Tests Cotemporary Calculus 9 Practice 2: (a) Put b. The b ( ) ( + ) 3 ( ) so L is positive ad fiite, ad ad b both coverge or both diverge. Sice we ow we ca coclude that (b) b b diverges (P test, p or as the Harmoic series), Put b The 4 diverges 4 4 so L is positive ad fiite, ad ad b both coverge or both diverge. Sice we ow b 2 coverges (P test, p2), we ca coclude that 4 coverges. 4 Practice 3: (a) which diverges (P test, p ) so diverges. (b) / /2 which coverges (P test, p 3/2) so coverges. (c) 2 26 which diverges (P test, p ) so diverges.

10 0. Positive Term Series: Compariso Tests Cotemporary Calculus 0 Appedix: Proof of the Limit Compariso Test (a) Suppose b coverges ad that Sice b L, there is a value N so that b L, a positive, fiite value. b < L + whe N. The < b. (L + ) whe N, ad a < (L+). N b. Sice b N N coverges, we ca coclude that a N coverges. (b) Suppose b diverges ad that Sice b L, there is a value N so that b L, a positive, fiite value. b > L/2 > 0 whe N. The > L 2. b whe N, ad a > L 2. N b. Sice b N N diverges, we ca coclude that a N diverges.