Ratio Test Recall that every convergent series X a k either

Transcription

1 Ratio Test Recall that every convergent series X a either X converges absolutely a converges, thus so does X a,or X converges conditionally a converges, but X a does not We will loo at two tests (Rato Test and Root Test) forabsoluteconvergenceofseriesofgeneralterms. They, of course, also apply to series of positive terms as a special case. Theorem (Ratio Test for series of general terms) Let X a be a series with non-zero terms and let L = If L <, then X a converges absolutely. If L > orl =, then X a diverges. a + a. Then Theorem (Ratio Test for series of positive terms) Let X a be a series with positive terms and let L = If L <, then X a converges. If L > orl =, then X a diverges. a +.Then a Math 67 (University of Calgary) Fall 05, Winter 06 / 6

2 Proof of the Ratio Test Proof of the Ratio Test: If L <, then consider a small open interval (A, B) containingl. We may assume that B <. Then for all large enough, a + a Then a + a < B =) a + < B a. is inside the interval (A, B). The above inequality holds for all large enough, say, for all larger than a certain integer N. Let a = a N+. It follows that a N+ < B a N+ = ab a N+3 < B a N+ < B(aB) =ab a N+4 < B a N+3 < B(aB )=ab 3... So, X a is ultimately bounded above by a geometric series with common ratio r = B <. Since X ab K converges, by the Comparison Test, X a converges, i.e., X a converges absolutely. AsimilarargumentshowsthatifL >, then X a is ultimately bounded below by a (divergent) geometric series with common ratio r >. Finally, all p-series X have L =. But some converge (p > ) and some diverge (0 < p apple ). p Math 67 (University of Calgary) Fall 05, Winter 06 / 6

3 Tip: If factorials appear, try the ratio test. Notice how cancellation between factorials simplifies the ratio in the examples below. Example Does ()! 4 converge? What about ( )!? Solution: The first series has all terms positive. The second is alternating. L = a + = a ((+))! 4 + ( +)! = ()! ()! = ( +)( +) ()! ()! ( +)( +) = 4 =. =) by the ratio test, ()! 4 diverges For the second series: L = = ( )( ) ( +)! ( +)( +) apple Recall or using l Hô pital s rule: a + a = ( )! ( ) + (+)! (+) + = ( ) =! ( ) + x = e = x! x e ( +) = ( ) + ( +)! ( +) + (+) = ( )! + < =) series converges absolutely Math 67 (University of Calgary) Fall 05, Winter 06 3 / 6

4 Root Test Theorem (Root Test for series of general terms) Let X a be a series with non-zero terms and let L = If L <, then X a converges absolutely. If L > orl =, then X a diverges. q a. Then Theorem (Root Test for series of positive terms) Let X a be a series with positive terms and let L = If L <, then X a converges. p a.then If L > orl =, then X a diverges. Proof: If L <, then for all large enough, p a < B for some B <. Then a < B =) X a is ultimately bounded above by a (convergent) geometric series =) X a converges. AsimilarargumentshowsthatifL >, then X a is ultimately bounded below by a (divergent) geometric series with common ratio r >. Finally, all p-series X apple p have L =. Recall p =. Math 67 (University of Calgary) Fall 05, Winter 06 4 / 6

5 Tip: If a has a -th power, try the root test. Example Does converge? Solution: The terms are ultimately positive (all positive except the first one). L = p a = 4 5 +! / 4 5 = + =>. By the root test, the series diverges. Ratio test and Root test: Could the other one wor if one of them is inconclusive? The short answer is no. It turns out that If a + a = L, then q a = L too. In particular, if the ratio test is inconclusive (L =),thentheroottestwillalsobeinconclusive. If the root test is inconclusive (L root =),theneithertheitoftheratiosdoesnotexist(andtheratio test is not applicable at all) or the ratio test is inconclusive too (L root = L ratio,bothbeingnow). Math 67 (University of Calgary) Fall 05, Winter 06 5 / 6

6 Further examples/exercises Determine whether the following series are divergent, absolutely convergent or conditionally convergent. ( 3)! (4 +)! 3 +3 ( +)!(3)! (ln ) 4 sin ( )! Math 67 (University of Calgary) Fall 05, Winter 06 6 / 6