0.1. Geometric Series Formula. This is in your book, but I thought it might be helpful to include here. If you have a geometric series

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1 Covergece tests These otes discuss a umer of tests for determiig whether a series coverges or 0.. Geometric Series Formula. This is i your oo, ut I thought it might e helpful to iclude here. If you have a geometric series X ar = a + ar + ar 2 + ar =0 with a 6= 0 the this coverges if ad oly if r <. If we do have r < thethegeometric series coverges to a. r Prolem 0. Does the series P = coverge? Justify your aswer ad compute its value. 3 Solutio: P = = is a geometric series with a = ad r =. Sice r <, we see that the series coverges to a = /3 = /3 = y the Geometric Series r /3 2/3 2 Formula Itegral test. Cosider the sum P = a.letf(x) eafuctiosuchthat: f(x) ispositiveaddecreasigforallx>0ad f() =a for all =, 2,...,. The: If R f(x)dx is fiite the P = a coverges. If R f(x)dx is ot fiite the P = a Prolem. Show that P = Solutio: Let f(x) =. Note that f(x)satisfiesthehypothesesoftheitegraltest,amely: x f(x) ispositiveaddecreasigforallx>0, ad f() = for all =, 2, 3,... We may thus apply the Itegral Test. We recall from Chapter II that R dx is ot fiite. x (If you had forgotte this, it comes from either the chart for improper itegrals, or from adirectcomputatio.) Sice R dx is ot fiite, the Itegral Test shows that the series P x = Prolem 2. Show that P = coverges. 2 Solutio: Let f(x) =. Note that f(x) satisfiesthehypothesesoftheitegraltest, x 2 amely: f(x) ispositiveaddecreasigforallx>0, ad f() = for all =, 2, 3,... 2 We may thus apply the Itegral Test. We recall from Chapter II that R dx is fiite. If x 2 you had forgotte this, it comes from either the chart for improper itegrals, or from the

2 direct computatio: Sice R Z Z dx =lim x2! x dx 2 =lim x! =lim! apple + =. dx is fiite, the Itegral Test shows that the series P x 2 = 2 coverges. No-example of itegral test! To use the itegral test, it is essetial that you verify all of the hypotheses. Watch what happes if you wor with a fuctio f(x) ethatisot positive ad decreasig. We cosider P = si( ) adf(x) = si( x). For the itegral we have: Z apple si( x)dx =lim! cos( x) which does ot exist. By cotrast: X si( ) = si( )+si(2 )+si(3 )+ = =0. = The Itegral Test fails i this case ecause f(x) isot a positive ad decreasig fuctio. Moral: always chec the hypotheses of ay test you are usig! 0.3. The test. The series P p = coverges if ad oly if p>. p This is t really a separate test. It is just a immediate cosequece of the itegral test, ad the fact that R dx coverges if ad oly if p>. x p 0.4. Term test. Cosider the series P = a. If lim! a 6=0the P = a However: if lim! a =0,thethetestisicoclusiveadyoumustusesomeother test. Prolem 3. Show that P 2 = Solutio: We will use the Term Test. We compute lim a!! ! ! = = 3 Sice the limit of the terms is 6=0,theTermTestshowsthattheseriesP 2 3 =

3 0.5. Ratio test. Cosider a series P = a.computethelimitigratio: L :! a + a If L<the P a coverges. If L>the P a If L =orifthelimitdoesotexist,thethetestisicoclusiveadyoumustuse some other test. Prolem 4. Show that P ( ) 3 coverges. 3 Solutio: We will use the Ratio Test. Compute L :! a + a! ( +) 3 /3 + 3 /3! ( +) 3 = Sice L = 3 <, the Ratio Test shows that the sum coverges. Prolem 6. Show that the Taylor series for e x coverges for ay fixed value x =. Solutio: If we tae T e x ad plug i x = we get: T e x x= = X =0! x x= =! We wat to use the Ratio Test. So we compute the limit: L! + /( +)! /! +! X =0!! ( +)!! Sice L =0<, the Ratio Test implies that this series coverges. + =0. Prolem 7. Show that P! Solutio: We will use the Ratio Test. For this, we compute the limitig ratio: L! ( +) + /( +)! /! ( +) +! ( +) ( +)! Sice L = e>theratiotestshowsthattheseries + = e.! Note that showig that the Taylor series T e x x= coverges is ot quite the same as showig that the sum coverges ad that the limit is e. We will discuss this i Chapter V.5.

4 No-example of Ratio Test. Cosider the series P =.Ifwetrytheratiotest,we 2 +5 get: L! (+) 2 +5(+) 2 +5! 2 +5 ( +) 2 +5( +) 2 +5! ! Sice the iggest term i sight is 2 we scale top ad ottom y to get: 2 / ! / ! = = Sice L =,theratiotestisicoclusive. Thus, to determie whether the series P = coverges or diverges, we would eed 2 +5 to use aother test. (Hit: the Limit Compariso Test or the Itegral Test.) 0.6. Root test. Cosider the sum P = a. Defie L := p a. (Note that there is a asolute value uder the square root!) If L<the P a coverges. If L>the P a If L =orifthelimitdoesotexist,thethetestisicoclusiveadyoumustuse some other test. Note: The root test is most useful whe the th term already has a th power i it. Prolem 9. Show that P = si ( )coverges. Solutio: Recall that si ( )= si( ). Sice there is already a th power i the th term, we will use the Root Test. For this, we compute q q L si ( ) si(!! ) si( ) = si(0) =0.! Sice L =0< theroottestshowsthatthisseriescoverges. Prolem 0. Show that P 3 = coverges. 2 Solutio: We wat to use the Root Test. For this, we must compute: s L 3 3! ! = 3.

5 Sice L = <, the Root Test implies that this series coverges Limit Compariso Test. This is very similar to the test we saw for improper itegrals. Let P =0 a ad P =0 e two series. Assume that a > 0ad > 0forall, ad a lim! exists ad equals a positive umer. (I particular, this limit caot equal 0or!) The either oth series coverge or oth series diverge. Prolem. Determie where P = a coverges, where a = 2 aswer Justify your Solutio: We will use the Limit Compariso Test ad the Itegral Test. As!,the largest P terms i the umerator ad deomiator will domiate. So we compare this to = where 2 =.Weotethat: 3 a > 0forall, siceoththeumeratoraddeomiatorallpositiveforall, ad > 0forall sice is positive for all. Also: a 2 lim!! ! = So the limit exists ad equals a positive umer. Thus we may apply the Limit Compariso Test to coclude that either oth series coverge or oth series divert. Next we use the Itegral Test to aalyze P =.Letf(x) =.Weote: x f(x) ispostivieaddecreasigforallx>0adthat f() = for all =, 2, 3,... Thus we may apply the Itegral Test. Sice R f(x)dx = R Test implies that P = the origial series, P = dx does ot exist, the Itegral x By the Limit Compariso Test, this i tur implies that: also Prolem 2. Show that P = a coverges whe a = 3 + p /2 +4 Hit: first use the Limit Compariso Test with = 3 to see that P = coverges. 4 2 Prolem to discuss: what test would you use? P si(/) P P 2 + e = 4 2.TheusetheItegralTest