Section 5.5. Infinite Series: The Ratio Test

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1 Differece Equatios to Differetial Equatios Sectio 5.5 Ifiite Series: The Ratio Test I the last sectio we saw that we could demostrate the covergece of a series a, where a 0 for all, by showig that a approaches 0 as as fast as the terms of aother series with oegative terms which is already kow to coverge. Both of the techiques developed i Sectio 5.4, the compariso test ad the limit compariso test, proved to be very useful; however, they both suffer from the drawback of requirig that we first fid a series of kow behavior which allows for the proper compariso with the series uder cosideratio. I this sectio we shall cosider aother test for covergece, the ratio test, which determies whether or ot the terms of a series are approachig 0 at a rate sufficiet for the series to coverge without referece to ay other series. Although this test does ot require kowledge of ay other series, it has the limitatio of beig icoclusive i certai circumstaces. Ufortuately, there is o sigle test for covergece which is useful uder all coditios. The ratio test determies if the terms of a give series are approachig 0 at a rate sufficiet for covergece by cosiderig the ratio betwee successive terms of the series. Specifically, suppose a > 0 for =, 2, 3,... ad a + lim = ρ. (5.5. a If ρ <, the there is a iteger N ad a umber r with ρ < r < such that a + a < r (5.5.2 for all > N. The for all > N, so a + < ra (5.5.3 a N+2 < ra N+, a N+3 < ra N+2 < r 2 a N+, a N+4 < ra N+3 < r 3 a N+, ad, i geeral, for ay iteger m > 2, a N+m < r m a N+. (5.5.4 That is, a N+m r m < an+ (5.5.5 Copyright c by Da Sloughter 2000

2 2 Ifiite Series: The Ratio Test Sectio 5.5 for m = 2, 3, 4,.... Lettig = N + m, i which case m = N, we have for all > N +. Thus a is O(r N. Moreover, a < an+ r N (5.5.6 r N = r N r (5.5.7 coverges sice r is a geometric series ad 0 < r <. Thus a coverges by the limit compariso test. Now suppose ρ >, i which we iclude the possibility that ρ =. The there is a iteger N such that a + > (5.5.8 a for all > N. Hece a + > a for all > N, ad so a N+ < a N+2 < a N+3 < a N+4 <. (5.5.9 I particular, a > a N+ for = N + 2, N + 3, N + 4,.... It follows that either lim a does ot exist or lim a > a N+ > 0. (5.5.0 Thus a diverges by the th term test for divergece. We ow summarize the above results. Ratio Test Suppose a > 0 for =, 2, 3,... ad The a coverges if ρ < ad diverges if ρ >. a + ρ. (5.5. a The examples below will show that the ratio test is icoclusive if ρ =. Namely, the third example cosiders a diverget series for which ρ = ad the fourth example cosiders a coverget series for which ρ =. Hece some other test will be ecessary to determie the behavior of ay series for which the ratio test yields ρ =. 3, a = 3,

3 Sectio 5.5 Ifiite Series: The Ratio Test 3 =, 2, 3,..., the a + ρ a Thus ρ < ad the series coverges. =, 2, 3,..., the a + ρ a Thus ρ > ad the series diverges. the ratio test yields For the harmoic series ρ + ( ( + 3 a = 5 + 2, 5 + 2, 3 + ( ( , + = ( 3 lim + = 3. + This shows that it is possible for a series to diverge whe ρ =. the ratio test yields ρ For the coverget p-series ( , 2 ( 2 ( This shows that is possible for a series to coverge whe ρ =. + 2 = 5 lim + 3 =. + 2 = 5. =.

4 4 Ifiite Series: The Ratio Test Sectio 5.5 =, 2, 3,..., the a + ρ a 3 + ( +! 3! 3!, a = 3!, ( (! 3 + ( +! = 0. Thus ρ < ad the series coverges. Problems. For each of the followig ifiite series, decide whether the series coverges or diverges ad explai your aswer. 2 (a 2 (b (c (d! ( +! (e (g 2!! (f (h 4 2 π For each of the followig ifiite series, decide whether the series coverges or diverges ad explai your aswer. 3 (a (b (c (e (g 3 +5 (2!!! (2! (d (f (h (2!!! (

5 Sectio 5.5 Ifiite Series: The Ratio Test 5 3. Defie f(x = x 2. (a Fid the domai of f. That is, fid all values of x for which the series coverges. (b Plot a approximatio to the graph of f o the domai foud i (a usig f(x x 2 00 x Defie g(t = =0 t 2!. (a Fid the domai of g. That is, fid all values of t for which the series coverges. (b Plot a approximatio to the graph of g o the iterval [ 2, 2] usig =0 g(t t 2! 50 =0 t 2!. 5. Suppose the terms of the series a satisfy the differece equatio a + = ( + a 2 with a = 0. Does this series coverge or diverge? Explai. 6. Suppose, for =, 2, 3,..., a 0 ad α a. Show that a coverges if α < ad diverges if α >. This result is kow as the root test.

62. Power series Definition 16. (Power series) Given a sequence {c n }, the series. c n x n = c 0 + c 1 x + c 2 x 2 + c 3 x 3 +

62. Power series Definition 16. (Power series) Given a sequence {c n }, the series. c n x n = c 0 + c 1 x + c 2 x 2 + c 3 x 3 + 62. Power series Defiitio 16. (Power series) Give a sequece {c }, the series c x = c 0 + c 1 x + c 2 x 2 + c 3 x 3 + is called a power series i the variable x. The umbers c are called the coefficiets of