11.6 Absolute Convergence and the Ratio and Root Tests

Transcription

1 .6 Absolute Covergece ad the Ratio ad Root Tests The most commo way to test for covergece is to igore ay positive or egative sigs i a series, ad simply test the correspodig series of positive terms. Does it seem reasoable that the covergece of the series = 3 = should say aythig about whether the modified series every third term is egative) coverges? The fact that it does is somewhat remarkable. Before we uderstad this, we eed to split the otio of covergece ito two cases. Defiitio. A series a is absolutely coverget if a coverges. A series a is coditioally coverget if it coverges but ot absolutely. Examples. The series ) is absolutely coverget, sice the p-series coverges. is coditioally coverget: it coverges, but the har-. The alteratig harmoic series ) moic series diverges. Theorem. If a series a is absolutely coverget the it is coverget. Proof. We use the fact that a = a + a ) a. It follows, for ay sequece a ), that 0 a + a a Assumig that a is absolutely coverget i.e. that a coverges), we may apply the compariso test to see that a + a ) coverges. It follows, by the series laws, that a = a + a ) a is coverget. For a give series a, we have show that exactly oe of three thigs must be true: a is absolutely coverget. a is coditioally coverget. a is diverget. The theorem allows us to apply ay of the tests we ve see that require oly positive terms to ay series. Such tests will oly be able to show absolute covergece or divergece.

2 Example Show that a = si ) coverges. Sice 0 si ), we observe that a hece a coverges by the compariso test. It follows that a coverges absolutely: i particular, a coverges. The Ratio Test The ratio test is perhaps the easiest of the covergece tests to use, but it is also oe of the most likely to be icoclusive. It is particular useful for decidig o the covegece of series cotaiig expoetial ad factorial terms. a Theorem Ratio Test). Let a be a series ad let L = + a If L < the a is absolutely coverget If L > the a is diverget If L = the the ratio test is icoclusive, if it exists. There are three possibilities: Sketch Proof. If L < the r = +L lies half way betwee L ad. Takig ɛ = L i the defiitio of it, we see that, for larger tha some fixed N, we have a + r a = a a N r N Now apply the compariso test to compare a a N r N r =N+ =N+ which is a coverget geometric series. It follows that a is absolutely coverget. If L >, the for sufficietly large we have a + a. It follows that the sequece a ) does ot coverge to zero, whece the th term/divergece test says that a diverges. Examples. The ratio test is useless for series of ratioal expressios, as the it will always be L =. Use compariso test istead. For example, we kow that coverges, ad diverges. The ratio test calculatios i each case are / / + ) = + ) = ad / / + ) =

3 . If a = )! the a + a = ) +! + )! ) = + = 0 It follows that a is absolutely coverget. 3. The ratio test is useful whe you wat to igore polyomials. For example, if a = 3 the a + a = ) + 3 = 3 + ) = 3 Note the irrelevace of the polyomial terms. Sice 3 > we coclude that a diverges. 4. This fial, tougher, example requires you to recall a it from earlier i your calculus career. If a =! the a + a = + )! [ ] + ) +! = = [ ] + + = e Sie e < we coclude that a is absolutely ) coverget. The Root Test The root test is very similar to the ratio test. I the abstract it is slightly more useful, although it is typically less applicable to cocrete series. Theorem Root Test). Let L = a = a /, if it exists. There are three possibilities, with the same coclusios as the ratio test: If L < the a is absolutely coverget If L > the a is diverget If L = the the root test is icoclusive Sketch Proof. If L < the a L for sufficietly large. We ow use the compariso test with the coverget geometric series L. If L >, the a = L >. I particular, the sequece a ) does ot coverge to zero, ad so the series diverges. Because the root test ivolves takig th roots, it is almost etirely useless! Uless a series is of the form b ), it is very ulikely that the root test will be at all useful. + x ) = e x. Sice each a is positive, absolute covergece is the same as covergece. 3

4 Examples. Cosider a =. Sice a / = ) / = = 0 ad 0 <, the root test quickly shows that is absolutely) coverget. This example ca also be easily doe by compariso: if, the = = = which coverges.. If a =, +) the we compute. a / = + = + = The root test shows that a is absolutely coverget. 3. Fially, aother example usig the it defiitio of e x. If a = ), the a / = ) = e < whece a is absolutely) coverget. L = e so coverget. Switch to + for couterexample. To illustrate the proof of the root test for L >, cosider modifyig the last example. If b = + ), the root test produces a it b / = e > whece b diverges. However, it should be obvious that b = + ) > = b = 0 so the series b diverges by the th term/divergece test: the root test was t eeded at all! Suggested problems. a) Explai the differece betwee absolute ad coditioal covergece. b) ) = coverges. Is the covergece absolute or coditioal? Explai your aswer.. Use the Ratio or Root test to decide whether the followig series coverge. 4

5 a) 3! = b) Harder) ) 3 = 3. a) Cosider the series = b) Cosider the followig sequece: { if odd a = 3 if eve i. Attempt to apply the root test to ii. Repeat a) for the ratio test. iii. Does!). Use the ratio test to decide whether this series coverges. )! a. What happes? = a coverge? Prove it ad, if it does, fid its value. = 5