Section 11.6 Absolute and Conditional Convergence, Root and Ratio Tests

Transcription

1 Sectio.6 Absolute ad Coditioal Covergece, Root ad Ratio Tests I this chapter we have see several examples of covergece tests that oly apply to series whose terms are oegative. I this sectio, we will lear a test that will i some cases allow us to use the same tools for series with positive ad egative terms. We begi by recordig a defiitio: Defiitio. A series a coverges absolutely if the series of absolute values a coverges. A series coverges coditioally if a coverges but a does ot coverge. We have already see that the alteratig harmoic series ( ) = = coverges, but the harmoic series ( ) = = = ( ) = = = does ot. Sice the alteratig harmoic series coverges, but the series we get whe we chage all of its terms to positives does ot, we say that the alteratig harmoic series coverges coditioally. O the other had, the series ( ) ( ) 3 = coverges absolutely sice the series of absolute values is geometric with r = 3 <. ( 3) ( ) = = ( 3 = ) It turs out that ay series that is absolutely coverget must be coverget itself: Absolute Covergece Test. If a coverges, the = a does as well. I other words, if the ew series we get from a by makig all of its terms positive is a coverget series, the the origial series coverges as well. This seems plausible it is more likely that a series with both positive ad egative terms will coverge (sice may of the terms will effectively cacel), so if the series with all positive terms coverges, so must the origial series. =

2 Sectio.6 There are several importat poits to ote here: first of all, the theorem is helpful sice we already have several tests for series that have positive terms. To test a series that does t have all positive terms, we ca take absolute values ad use oe of the previously studied covergece tests that apply to series with positive terms; if the series of absolute values coverges, the the origial series does too. If the series is ot absolutely coverget, it may still be coditioally coverget, or it may diverge after all. To check, try the Alteratig Series Test (if the origial series was alteratig) or the th term test. Example. Determie if each of the followig series coverges absolutely, coditioally, or diverges. ( ) ta( ) = To determie if the series coverges absolutely, we eed to determie if ta( ( ) ) = ta( ) = ta( ) = = coverges or diverges. Let s try the it compariso test. For large, 0 so that ta( ) 0. It seems that the fuctio ta( ) behaves similarly to, so we will replace the umerator with. We compare = ta( ) to = 3 : ta( ) 3 3 ta( ) ta( ). Notice that the last it above has idetermiate form 0 ; thus we rewrite it so that L Hopital s Rule applies: ta( ) ta( ) LR = sec ( ) sec ( ) = sec 0 =.

3 Sectio.6 The compariso series quotiet of the terms is, 3 coverges sice it is a p-series with p = 3; sice the it of the = ta( ) coverges as well by the it compariso test. Sice the series of absolute values coverges (i.e. is the origial series coverges absolutely), we coclude that ( ) ta( ) = coverges as well, this time by the absolute compariso test. Example. Determie whether the series ( ) = coverges absolutely, coverges coditioally, or diverges. It is easy to see that the series does ot coverge absolutely by the th term test: ( ) =. Sice the it of the terms is o-zero, the series of absolute values diverges. The origial series may still coverge (i.e., the series might coverge coditioally), or it may diverge; however, the result of the previous test idicate that we should try the th term test o the origial terms. If is eve, we already kow that ( ) =. 3

4 Sectio.6 If is odd, the ( ) =. Thus the geeral it does ot exist, so that the series fails the th term test, ad diverges. Example. Determie whether the series ( ) l = coverges absolutely, coverges coditioally, or diverges. Let s start by testig for absolute covergece; we eed to look at the series ( ) l = = We kow that we ca itegrate l = l = = l. by u-substitutio, so we try the itegral test. Settig u = l, so that du = d, we have l d = u du = l u + C = l(l ) + C. So l b d b l d b (l(l ) b ) b (l(l b) l(l )) =. 4

5 Sectio.6 Sice the improper itegral diverges, the series of absolute values does too. Thus the origial series is ot absolutely coverget, but might still be coditioally coverget. To check, let s try usig the Alteratig Series Test. We have each term a is oegative, ad it is clear that Fially, we ote that ( + ) l( + ) < l Sice the it is 0, the origial series a = l = l ; sice ( + ) l( + ) > l. l = 0. ( ) l = coverges by the alteratig seres test. Sice it does ot coverge absolutely, we specify that the series coverges coditioally. Root ad Ratio Tests The Ratio Test is a covergece test that ca be particularly helpful whe the series at had has terms ivolvig factorials or th powers: Ratio Test. Let a be a series so that a + a = ρ.. If ρ > or ρ =, the series a diverges.. If ρ <, the series a coverges absolutely (i.e., a ad a coverge). 3. If ρ =, the test is icoclusive. The Root Test is geerally helpful if terms of the series ivolve th powers; it is ot usually helpful whe there are factorials ivolved. Root Test. Let a be a series, ad suppose that a = ρ. 5

6 Sectio.6. If ρ > or if ρ =, the series a diverges.. If ρ <, the series coverges absolutely. 3. If ρ =, the test is icoclusive. Example. Determie whether the sereies coverges or diverges. ( 3 = The th power i the fractio above idicates that the root test could be a good way to proceed. As we evaluate the it, recall that / =. ) ( 3 ) = 3 = 3. ( ) 3 Sice the it is greater tha, the series diverges by the root test. Notice that the itegral test would ot be helpful i this example, sice we do ot kow how to itegrate a fuctio of the form f(x) = x( 3 )x. Example. Determie whether the series coverges or diverges.! e = The terms of the series have a factorial i them, so the ratio test seems like a good place to start. Sice all of the terms are positive, we ca igore the absolute value sigs, ad evaluate the it 6

7 Sectio.6 a + a (+)! e (+)! e ( + )!e.!e (+) Now the fractio above may be rewritte; sice ( + )!! = ( + ) ( )... ( )... = +, we have ( + )!e!e (+) ( + )e e ++ ( + ) LR = = 0 e ++ ( + ) e + e + sice e+ =. Sice the it is 0, the series coverges by the ratio test. Example. Determie if the series coverges or diverges. =! It seems reasoable to guess that either the root test or the ratio test will be appropriate for this example. Notice that if we try the root test, we will eed to hadle terms of the form!, which will ot readily simplify. Thus our best choice here is probably the ratio test. We eed to evaluate 7

8 Sectio.6 a + a (+)! (+) +! ( + )!!( + ) + ( + ) ( + ) + ( + ) ( ). + The it looks ufathomable, but a simple trick will help us evaluate it: ( ) + ( + ) ( + ) = e sice ( + = e, ) as we saw i sectio.. Sice e >, we kow that e < ; so the series coverges by the ratio test. 8