n a n = L, then lim cos(2/n) = cos(0), lim arctan 2n = π/2, lim = limit does not exist. lim

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1 Usig L Hospitals Rule fid the followig its e 0, 0, si(π/ ) π. (l ) If f(x) is a cotiuous fuctio, ad a L, the f(a ) f(l). Usig this fid the followig its ,. e/ e 0, To Determie the it of fractios like We take the 4 π ta( +8 ) ta(π/4), + cos(/) cos(0), arcta π/, biggest thig i umerator biggest thig i the deomiator (l ) 0(By LHospital s Rule) , e +e e +l,. For example e it does ot exist. +e e +l e e /, (l ) + (l ) +. (l ) + (l ) + Remember/Recall the hierarchy of bigess amog differet kid of fuctios. logarithmic fuctios(l, (l ), etc) <<,,etc << polyomials << expoetials(e,, etc) << factorials(!,(-)!, etc). +5 usig this fid the followig its ) l( ++) l, e +e e 5 + (/5) , + 5, + diverges, + 5 (Solutio ) 0, + + e + π π l( + Squeeze Theorem If {a }, {b }, {c } are three sequeces such that a b c, ad a c L. The b L exists ad is equal to L. 5 ( + 5 ) cos Usig squeeze theorem fid the followig its si 4 0, + ( ) 0,! 0 Fid + 5, e + + Sol + 5 ( 5 ) 0, e + e + 9 ( e ) diverges Let {a } be sequece. The a 0 if ad oly if a 0. Usig this determie if the it i the followig cases is 0. ( ) cos(/) diverges, ( ) 0, e 0 + ( ) arcta() diverges, ( ) ( ++) 0, ( ) + Let {a } be a alteratig sequece. for example {( ) ( ) + }, { cos(π)}, {! }, { ( ) + }, that is a ( ) b or a ( ) b, where b > 0. The a b, ad

2 a exists if ad oly if a b 0. If a does coverge the we must have a 0. If a 0, the a diverges. This is oly true for alteratig Sequeces Mootoic Sequece Theorem Let {a }, be bouded i.e m a M for some umbers m, M. ad {a } is evetually mootoic that is for some umber N we have a + > a for all > N or a + < a for all > N. that is evetually icreasig or evetually decreasig. The a coverges ad satisfies m a M. usig MST fid the followig its, suppose a, a + a, ad assume (doot prove) this sequece is icreasig ad a < for all, deduce a coverges ad fid the it. As + 5 suppose a, a + a, ad assume (doot prove) this sequece is decreasig ad 0 a for all, deduce a coverges ad fid the it. As 5 Suppose {a }, {b } are two sequeces, ad a coverges, ad b coverges. The (a b ), ad a + b coverge. ad (a b ) ( a )( b ), ad (a + b ) a + b. For example (cos(/) arcta()) π/, (e/ + e ) + 0 Suppose {a }, {b } are two sequeces, such that a coverges ad a 0, but b diverges. The (a b ), ad a + b diverges. For example cos() + diverges because + 0 coverges, but cos() diverges. Usig this determie if the followig its coverge cos() + its coverge. diverges, cos + arcta() si(/) 0, 0, e e. WATCH OUT the last two The followig is a flowchart/steps oe ca use to determie the covergece of a series

3 start s is give or ca fid s i.e telescopig series a is a geometric series fid s s diverges, the a diverges fid a by pluggig i the startig value of ito the formula, ad r r > the a diverges s L coverges, a coverges ad L r <, the a coverges ad a r TOD:is a 0? if a 0, TOD implies a diverges a+ a >, the a diverges a 0, TOD is icoclusive, apply ratio test, fid a+ a a+ a <, the a coverges absolutely a+ a, the Ratio Test is icoclusive, Is this a alteratig series if b 0, the is {b } evetually decreasig o If this is a alteratig series, is b 0 b 0, ad {b } is evetually decreasig, the AST implies a coverges b 0, the ( ) b diverges if a approximatio to ( a) b is reqd, the fid such that b + maxerror, the by ASET s is a good eough approximatio is a a positive series, i.e is a 0 for all is a coverget(use Comparisio test or it Comparisio test) a coverges, the a coverges absolutely, hece coverges a diverges, the a may still coverge Apply Limit Comparisio Test LCT doesot help, the use Compariso Test try itegral Test

4 4 Test of Divergece(TOD): Let a be a series. If a 0, the a diverges. if a 0, the the test is icoclusive. This is slightly differet from the book, but is easier to apply i the cases like Usig TOD, determie if the followig series coverge. diverges, diverges, diverges, arcta() diverges, (+) (+) diverges, ( ) ( ) + cos() diverges, cos(/) diverges, ( ) + diverges, ( ) ++ icoclusive + A geometric series a + ar + ar a + coverges if r <, the the sum is r, it diverges if r. a is called the first term, r is called the commo ratio. For the series the startig value of is, pluggig this ito the formula we get the first term a 74 0, ad r 7/0. sice r < we get sum is (74 /0 ) (7/0). while i the geometric series, we have r / >, hece this series diverges. + Determie if the followig geometric series coverge. If they do, the fid their sum. 0 ( 5), ( 9), + (, 4++9/4+7/6+, , ) , 6(0.9), cos() k. Let a, b be two series the if oe of the two series coverges, while the other oe diverges the (a + b ) diverges. If both the series co- verge, the (a + b ) coverges. If both the series diverges, we doot kow what happes to (a + b ). I class, we deoted this by C + C C, C + D D, D + D? usig this determie if the followig series coverge diverges, + / +, + [(0.8) (0.) ](fid the sum, leave your aswer as a fractio), ( e + (+) ) coverges, ( + 5 ) diverges Fid all values of c for which the series ( c oly for c, diverges for all other values of c. + ), coverges.as- Coverges

5 (v.importat) Give a series a, s is the sum of the terms from the start upto ad icludig the term oe gets by plugig i ito the formula of the series. For example for the series l, s / l + / l. if the series starts at, i.e a, the a s but for >, a s s. for example if the series is a s s /4 / /6. a coverges if ad oly if a a, ad it is give s +, the a s 0, while s coverges. ad if they do coverge the s. For example i the above example hece a coverges ad equals. Suppose the th partial sum of the series a /, a. determie if a. s a is s The fid a coverges yes. If it does fid the value of (v.importat) Usig the telescopig series method fid s for the followig series. the usig s determie if the series coverges. if it does the fid the value of the sum. ( + ) coverges, +4+ 7/ coverges, (+) coverges, l( + ) diverges. Itegral Test, Comparisio Test, Limit Comparisio Test, ca be used oly for series with positive terms. if a b, ad b coverges, it is NOT always true that a coverges. for example cosider the series ( ) < ( ). we caot use comparisio test because a is ot positive. the p-series, coverges if p > diverges if p. p. diverges as p.. usig the p-series test determie if the followig series coverge diverges, (.4 +. ) coverges, For example 5 diverges, typo, + + +

6 6 coverges Usig the itegral test determie if the followig series coverge e coverges, l. diverges Give a series like a e ++ e +. to apply it compariso test, to this we eed a secod series b, which is easier to hadle but such that ay ozero umber. Take b The it is clear. we kow coverges, hece the give series must coverge by it compariso test. a b or biggest summad i the umerator of a biggest summad i the deomiator of a e e. e ++ e + usig the it compariso test determie if the followig series coverge D, + C, 4.5 the th term, the use LCT, +4 D, + /4 + /7 + /0 + D (fid a formula for +9 C, + / + /5 + /7 + /9 + / + D, Give a series a, if the series coverges absolutely. I this case 4 + C, +4 + D, 4 C, C, + 4+ D, si( ) D a coverges, the we say a a must also coverge. Hece absolute covergece implies covergece. But the coverse is false. it may happe that a series coverges, but it does ot coverge absolutely. If the terms of a series are ot positive. The we caot apply compariso, LCT or itegral test to it. I this cases sometimes it helps to check the series for absolute covergece. to do that we check that the series a is coverget. if it is so, the a is absolutely coverget, hece coverget. But if Sera is ot absolutely coverget, it may still be simply coverget. For example coverget, but ot absolutely coverget. usig absolute covergece show that the series si() is coverget. ( ) is

7 7 i usig compariso test betwee two positive series, if the Bigger series coverges, the the smaller series coverges. If the smaller series diverges, the the bigger series diverges. (importat)to determie covergece of the series si(4) 4, we caot use compariso test directly to compare the series with 4, because si(4) 4 is ot positive for all. istead we use a combiatio of absolute covergece ad compariso test as follows si(4) 4 < 4, sice 4 coverges, we must have si(4) 4 coverges by compariso test. Hece the give series coverges absolutely, hece must also coverge. Use compariso test, or a combiatio of absolute covergece ad compariso si(4 test, if required, to determie if the followig series coverge. ) 4 absolutely cos Coverges, + Absolutely C., +( ) +si() Absolutely C., 0 Absolutely C., si Absolutely C. Alteratig Series Test Let b > 0 ad ( ) b be a alteratig series. The if the sequece of absolute values {b } is evetually decreasig, that is decreasig for > N, for some umber N, ad b 0. The the alter- atig series coverges. If b 0, The by TOD the alteratig series diverges. If the sequece {b } is ot evetually decreasig the the alteratig series Test is icoclusive. Usig the AST determie if the followig series coverges. l(+4) C, D, ( ) e C, ( ) ( ) ( ) ( ) C, + D, ( ) C, ( ) cos(π/) + arcta 8 C, ( ) C. To check that the sequece {b } is evetually decreasig, we have two methods. i first case it might be obvious that, it is so. for example i the case ( ). +. i these cases it is eough to just say so. If it is ot obvious for example i ( ) we use the first derivative test i the iterval (, ) as follows. Let f(x) x x +. The f (x) x( x ) (x +). i the iterval (, ) the factor x/(x + ) > 0, but ( x ) < 0 for x >. hece for x >, f (x) < 0. hece the sequece {b f()} is decreasig for.

8 8 Prove that the sequeces { e }, { }, are evetually decreasig. (Very Importat) Alteratig Series Estimatio Theorem Let ( ) b be a alteratig series which satisfies the coditios of the alteratig series Test, ad hece coverges. Let s ( ) b. Let s be the partial sum upto ad icludig the th term ( ) b, The the error i approximatig s by s s s b +. For example let s. The s ( /+/) /4. ( ) Usig the ASET, for each of the followig series fid such that, s s.00. ( ), ( ) l, ( ) 8, ( ), ( ) 6!. Ratio Test is geerally easier to apply tha the alteratig series Test. for example i the cases ( ) e, ( ) 8, checkig that the {b } is evetually decreasig is cumbersome, while the ratio test is straight forward. while applyig ratio test, we have to simplify ratios like (+)! () () (+) (+) (+)(+), ()! +, ((+))! (+) (+) (+) (+) (+)(+)! (+) (+)! ((+)+)! (+)! (+)! Give a series a, a the term we get whe we substitute ito the formula of the series. For example i the series 0 ( ) x + ( + )!, a ( ) x.+ (. + )!. see that a is the 4th term from the start. Give a series a, to apply ratio test to this series, we first fid a+ a ad simplify it.. The we take it s a+ a. If a+ a <, the series coverges absolutely. If a+ a >, the series diverges. If a+ a, the the Ratio test is icoclusive. Whe it is icoclusive, we should try AST, LCT, CT, if the series has egative terms the try a combiatio of absolute covergece ad compariso test, IT, i that order. Apply ratio test to the followig series. ( ) Absolutely C.,! D(is a hard questio, do t worry if you ca t do this), ( ) icoclusive, ( ) ( ) Absolutely C.,! Absolutely C.,! D, 00 D, k k Absolutely C., ( ) 0 k D, 4 (+)4 Absolutely C., ( )! + ()! Absolutely 0

9 C., + Icoclusive A power series is a fuctio which is represeted as a series. the series has a variable x i it. for some values of x, the power series coverges, while for the remaiig values of x,the power series diverges. The set of values of x for which the power series coverges is always a iterval of oe of the followig types ( ),( ], [ ], [ ). It is called the iterval of covergece. ofte deoted by IOC. the iterval of covergece is also the domai of the fuctio represeted by the give power series. the radius of this iterval is called the radius of covergece. ofte deoted by R. the ceter of this iterval is called the ceter of covergece. ofte deoted by a. Every power series coverges i it s ceter. So the iterval of covergece is ever empty. the iterval of covergece is always symmetric about the ceter. Suppose the iterval of covergece is (, ], the the ceter is ( + )/ /, while the radius is ( )/ /. the power i the word power series comes from the fact that the terms of the series cotai a power of x a, where a is the ceter of the power series. 9 A power series coverges absolutely i every ope iterval iside the iterval of covergece. For example the power series has iterval of covergece 0 (x ) [0, ). It coverges absolutely i (0, ). but at the boudary poit 0, it doesot coverge absolutely. Outside the iterval of covergece, a power series diverges. by the way the ceter of C. of this series is, ad the radius of C. of this series is. 0 Let c (x a), be a power series. the there are three types of radius of covergece. if the IOC is [a, a]. the radius is 0. i this case the power series coverges oly at it s ceter. for example 0!(x ) has IOC[, ]. if the iterval is of the type (a R, a + R], [a R, a + R], [a R, a + R), (a R, a + R), for some ozero fiite R, for example for the series (x ), which has IOC(0, ), hece has radius. The third type is whe the IOC(, ), i which case we say the radius of covergece is for example i the case (x )!. Fid the radius of covergece, ad the iterval of covergece of the followig x power series. ( ) +,R, IOC[-,), x + R, IOC(-,], (x) (x) R, IOC(-,), (0x)! R IOC(, ), R/0,IOC[-/0,/0], 0 ( ) x ( ) R,IOC (-,), x ()!x ()! R IOC(, ), R0 0 0 ( ) IOC[0,0], (x ) + R IOC(,4], (x ) R IOC[-/,5/), 0 0!(x )

10 0 R0 IOC[/,/], x 5 ( ) R IOC(, ). Very Importat for exam {c 0, c,...} is a sequece such that c 5 coverges but 0 c ( ) 8 diverges. what ca you say about the covergece of the followig series? 0 c Absolutely C., 0 0 c ( ) Absolutely C., 0 0 c 6 may or mayot coverge, c ( ) 0 Diverges, c Absolutely C., c ( ) 7 may or mayot coverge, 0 0 c x. Absolutely C. for x i ( 5, 5), Diverges for 0 x > 8. coverges at x 5, diverges at x 8.