10.1 Sequences. n term. We will deal a. a n or a n n. ( 1) n ( 1) n 1 2 ( 1) a =, 0 0,,,,, ln n. n an 2. n term.
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1 0. Sequeces A sequece is a list of umbers writte i a defiite order: a, a,, a, a is called the first term, a is the secod term, ad i geeral eclusively with ifiite sequeces ad so each term Notatio: the sequece a, a,, a, is also deoted by a or a a is the a has a successor Eample : Some sequeces ca be defied by givig a formula for the a =,,,, 3 th term. We will deal a. th term. cos a =cos,,,,( ), c. 3 a = 3,, 3,, 3, d. ( ) ( ) ( ) a =, 0 0,,,,, Eample : The Fiboacci sequece f is defied recursively by the coditios f, f, f f f for 3 Write dow the first 8 terms. Solutio:,,,3,5,8,3,, l Eample 3: Give f( ). Let a ( ) f, write dow the sequece a l l l Solutio: f( ) f( ) a f ( ). Thus a l
2 We wat to thik about graphig a sequece. To graph the sequece a we plot the poits ( a, ) as rages over all possible values o a graph. For istace, let s graph the sequece This graph leads us to a importat idea about sequeces. Notice that as icreases the sequece terms i our sequece, i this case, get closer ad closer to. We the say that is the limit of the sequece ad write Defiitio: A sequece a lim a lim has the limit L ad we write lim a L or a L as If we ca make the terms a as close to L as we like by takig sufficietly large. If lim a eists, we say the sequece Properties: a is coverget. Otherwise, we say the sequece a is diverget.
3 Eample 4: Fid lim Solutio: lim lim lim limlim 0 Eample: Determie whether the sequece cos is coverget or diverget. Solutio: If we write out the terms of cos, we have,,,,,, Clearly, the sequece limit does ot eist, so it is diverget. Theorem : Give the sequece a if we have a fuctio f( ) such that f ( ) a ad lim f ( ) L, the lim a L. Eample 5: Fid l lim Solutio: Let l l f ( ), the we have f ( ) ad l lim f( ) lim (usig L'Hospital's Rule) lim lim 0 Thus, we have Questio: Fid l l lim lim 0 lim e Squeeze Theorem for Sequeces: If a b c for 0 ad lim a lim c L, the limb L Theorem : if lim a 0, the lim a 0 Eample 6: Fid ( ) lim
4 Solutio: ( ) lim lim 0, thus, by Theorem, we have ( ) lim 0 Remark: We DO NOT have the followig: if lim a L 0, the lim a L (ever use this oe sice it is ot correct) Couter-eample: lim ( ) lim, but lim ( ) does ot eist. Theorem 3: The sequece r is coverget whe r ad diverget for all other values of r. lim r 0, r, r Eample 7: Determie the followig sequeces coverge or diverge. If the sequece coverges determie its limit. ( )! 4 9 c. l(4 3 00) l( 3 0 ) Solutio: ( ) 4 0!! 3 ( ) ( ) 3 ( ) 3 ( ) sice 3 ( ) 4 ad we kow that lim 0 ad lim 0 0, by Squeeze Theorem, we have ( ) by Theorem, we have lim 0! ( ) lim 0. Thus,! sice 6 r, by theorem 3, we have is diverget.
5 c liml l l lim[l(4 00) 3 l( 0 )] liml Questio: Eample: Determie the followig sequeces coverge or diverge. If the sequece coverges determie its limit. c.!! 4 9 Aswer:! 3 ( ) 3 ( ) sice 3 ( ) So is diverget.! c.! is coverget ad limit is 0 (please make up the details) , 4 4 r by theorem 3, we have 9 9 is coverget ad limit is 0 or 4 lim 0. 9 Defiitio: A sequece A sequece a is called icreasig if a a for all, that is a a a3. a is called decreasig if a a for all, that is a a a3. A sequece is mootoic if it is either icreasig or decreasig.
6 Eample 8: Show that a is decreasig. Solutio: We eed to show that a a, that is This is equivalet to ( ) ( ) ( )( ) [( ) ] 3 3 Sice, so is always true. Therefore, a a, ad so a is decreasig. a Defiitio: A sequece is bouded if there is a positive umber M, such that a M for all Mootoic Sequece Theorem: Every bouded, mootoic sequece is coverget. Eample: A sequece a is give by a, a a By iductio or otherwise, show that a is icreasig ad a bouded, more specific, that a 3. Deduce that a is coverget. Fid lim a Solutio: Let P be the statemet that a a ad a 3. Step : Whe. Clearly, we have a a 9 3 Step : We assume P is true, i other words, a a ad a 3 are true. Now we eed to show that P is true, i other words, a aad a 3 are true. a a a a a a a a which is the iductio hypothesis. a 3 a 3 a 9 a 7 which is true sice we assume that a 3.
7 So, P is true for all, i other words, a is icreasig ad a bouded, more specific, that a 3. By the Mootoic Sequece Theorem, the sequece a is coverget. If L lim a, the lim a L also. Sice a a lim a lim a L L L L 0 ( L )( L ) 0 Thus, L. (sice the limit L ca t be egative)
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