Strategy for Testing Series

Transcription

1 Strategy for Testig Series We ow have several ways of testig a series for covergece or divergece; the problem is to decide which test to use o which series. I this respect testig series is similar to itegratig fuctios. Agai there are o hard ad fast rules about which test to apply to a give series, but you may fid the followig advice of some use. It is ot wise to apply a list of the tests i a specific order util oe fially works. That would be a waste of time ad effort. Istead, as with itegratio, the mai strategy is to classify the series accordig to its form.. If the series is of the form p, it is a p-series, which we kow to be coverget if p ad diverget if p.. If the series has the form ar or ar, it is a geometric series, which coverges if r ad diverges if r. Some prelimiary algebraic maipulatio may be required to brig the series ito this form. 3. If the series has a form that is similar to a p-series or a geometric series, the oe of the compariso tests should be cosidered. I particular, if a is a ratioal fuctio or algebraic fuctio of (ivolvig roots of polyomials), the the series should be compared with a p-series. (The value of p should be chose as i Sectio 8.3 by keepig oly the highest powers of i the umerator ad deomiator.) The compariso tests apply oly to series with positive terms, but if a has some egative terms, the we ca apply the Compariso Test to a ad test for absolute covergece.. If you ca see at a glace that lim l a 0, the the Test for Divergece should be used. 5. If the series is of the form b or b, the the Alteratig Series Test is a obvious possibility. 6. Series that ivolve factorials or other products (icludig a costat raised to the th power) are ofte coveietly tested usig the Ratio Test. Bear i mid that a a l as l for all p-series ad therefore all ratioal or algebraic fuctios of. Thus, the Ratio Test should ot be used for such series. 7. If a f, where x f x dx is easily evaluated, the the Itegral Test is effective (assumig the hypotheses of this test are satisfied). I the followig examples we do t work out all the details but simply idicate which tests should be used. EXAMPLE Sice a l 0 as l, we should use the Test for Divergece. EXAMPLE s Sice a is a algebraic fuctio of, we compare the give series with a p-series. The compariso series for the Limit Compariso Test is b, where b s

2 STRATEGY FOR TESTING SERIES EXAMPLE 3 Sice the itegral x dx is easily evaluated, we use the Itegral Test. The Ratio Test xex also works. EXAMPLE Sice the series is alteratig, we use the Alteratig Series Test. EXAMPLE 5 e 3 k k k! Sice the series ivolves k!, we use the Ratio Test. EXAMPLE 6 3 Sice the series is closely related to the geometric series 3, we use the Compariso Test. Exercises 3 Test the series for covergece or divergece sl 9. k e k 0... l A k 0 6. Click here for aswers S k Click here for solutios. cos k k! k! e 3 si l 0. s.. 3. ta e 7. k l k 8. k 3 k 9. ta 30. s k k k 33. si 3. s s k 5 k cos 5 e j j s 3 5 l l sj j 5 (s )

3 STRATEGY FOR TESTING SERIES 3 Aswers S Click here for solutios.. D 3. C 5. C 7. D 9. C. C 3. C 5. C 7. D 9. C. C 3. D 5. C 7. C 9. C 3. D 33. C

4 STRATEGY FOR TESTING SERIES Solutios: Strategy for Testig Series. lim a = lim + = lim / +/ =6= 0,sotheseries X diverges by the Test for + = Divergece < for all, so P = coverges because p =>. 5. lim a+ a ( 3)+ 3 3(+) ( 3) + ( 3) + is absolutely coverget by the Ratio Test. 3 = dx x l x = lim t dx x l x = lim t + coverges by the Compariso Test with P =,ap-series that 3 = 3 <, sotheseries Let f(x) = x.thef is positive, cotiuous, ad decreasig o [, ), so we ca apply the Itegral Test. l x Z " # u =lx, Z Sice x l x dx = u / du =u / + C = l x + C, wefid du = dx/x Z Z t h i t ³ l x = lim l t l =. Sicetheitegral t 9. diverges, the give series a k = l diverges. k e k k =. Usig the Ratio Test, we get ek k= k= " µ # lim ak+ (k +) ek k + e k+ k = k e e = <, so the series coverges. e. b = > 0 for, {b} is decreasig, ad lim l b =0,sothegiveseries P ( ) + coverges by = l the Alteratig Series Test. 3. lim a+ a = 5. lim 3+ ( +) 3( +) ( +)! 3 ( +) 3 coverges by the Ratio Test. a + a = lim ( +)! 5 8 (3 +)[3( +)+] + = lim 3 +5 = 3 < coverges by the Ratio Test. 5 8 (3 +) so the series =0 7. lim / = 0 =,so lim ( ) / does ot exist ad the series Test for Divergece. + =3 lim =0<, sotheseries 5 8 (3 +) ( ) / diverges by the =

5 STRATEGY FOR TESTING SERIES 5 9. Let f(x) = l x. Thef 0 (x) = l x < 0 whe l x>or x>e,so l is decreasig for >e. x x 3/ l By l Hospital s Rule, lim = lim / / ( ) = lim =0,sotheseries = ( ) l coverges by the Alteratig Series Test. P ( ) µ P. = p. lim a = lim =0<, so the give series is absolutely coverget by the = = Root Test. µ 3. Usig the Limit Compariso Test with a =ta ad b =,wehave a ta(/) ta(/x) H sec (/x) ( /x ) lim = lim = lim = lim = lim b / x /x x /x x sec (/x) = => 0. Sice P = b is the diverget harmoic series, P = a is also diverget. 5. Use the Ratio Test. lim 7. = Z e coverges. ad sice a+ a l x dx = lim l x x t x t x ( +)! e (+) e ( +) e e ++ (usig itegratio by parts) H =. So P k l k (k +) 3 < k l k = l k k 3 k, the give series P k= 9. 0 < ta < π/ 3/. P 3/ = P = ta 3/ 3. lim a k = lim Thus, k= π/ 3/ = π P = coverges by the Compariso Test. 3 k + k =[divide by k ] lim diverges by the Test for Divergece. 3 k +k = l + = lim =0<,so e+ coverges by the Itegral Test, k l k 3 coverges by the Compariso Test. (k +) which is a coverget p-series (p = 3 > ), so 3/ (5/) k = sice lim (3/) k + µ Let a = si(/) ad b = a si(/).the lim = lim => 0,so b / limit compariso with the coverget p-series (p =3/ > ). 3/ = k =0ad lim = si(/) µ k 5 =. coverges by