Math 21, Winter 2018 Schaeffer/Solis Stanford University Solutions for 20 series from Lecture 16 notes (Schaeffer)

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1 Math, Witer 08 Schaeffer/Solis Staford Uiversity Solutios for 0 series from Lecture 6 otes (Schaeffer) a. r 4 +3 The series has algebraic terms (polyomials, ratioal fuctios, ad radicals, oly), so the test to try is limit compariso with a p-series: r r / so sice P coverges (by the p-series test), our series coverges as well. 3/ Because the series has oly positive terms, it i fact coverges absolutely. b. 5 e Several tests work here, but your first move should be to simplify the terms somewhat otably, the egative power is somewhat aoyig: e e 5 5 c. which is a geometric series with r e/. Sice e>, our r has absolute value, so this geometric series diverges. You ca also show this series diverges by the divergece test or by the ratio test (your L will come out to e/ > ). l This is a rare istace where you wat to use the itegral test. While geerally we discourage usig the itegral test (favorig limit compariso, ratio, ad alteratig series most of the time), this series requires it. The fuctio f(x) is positive, decreasig, ad cotiuous, so we ca apply the itegral test. Our series therefore has the same covergece/divergece behavior as the x l x itegral Z x l x R To itegrate this, we use u lxad du to get x Z x l x x l x lim [l l b l l ] b! ad therefore the itegral ad the series both diverge. llx + C so

2 d. e. f. ( ) ( +) 4 ( ) Simplifyig the terms of the series yields P, which is a geometric series with r /4, 4 so it coverges (absolutely). I this case, it coverges to /4. 3/4 l Series with logarithms either ivolve the itegral test (with u lx, as i c.) or the limit compariso test, buildig o the fact that l p for ay positive power p. p Here, we wat to use somethig like l (which is the statemet above with p /). The l p. 3/ Sice P coverges by the p-series test, ad domiates the terms of our series 3/ 3/ asymptotically, our series coverges as well. (It does so absolutely because its terms are all 0.) ! The expoetial ad factorial terms should alert you to use the ratio test:! a + a ! lim lim! ( +)! g. h. so sice L<, the series coverges absolutely. + P NOTE: You caot split this up as P because these series do ot coverge, + idividually (see Theorem 9.. i the book). Roughly speakig (ad perhaps disturbig): The commutative property of additio a + b b + a is ot actually guarateed whe sums are ifiite. Istead: Simplify the terms. This yields ( +) ( +) ( +) which coverges (absolutely) by limit compariso with a p-series, amely P. (It actually coverges to. If you write out the terms of the series you ca see why!) l Our two mai optios whe faced with a series cotaiig a logarithm is to try the limit compariso test usig what we kow about logarithmic growth, or the itegral test. I this

3 i. j. case, we see that the l term oly makes the deomiators of the terms grow more quickly tha those of. That is, l so sice P 0 si! coverges, our series also coverges (absolutely). I additio to logs, aother weird thig that pops up i series are sies ad cosies. There are two cases: If the sie/cosie s iput cotais, test values to see what the behavior is. Otherwise, i.e. if you just have si or cos, use the absolute covergece test ad/or the direct compariso test, startig from apple si, cos apple. NOTE: You caot apply direct compariso directly(!) to the series above, because it has egative as well as positive terms (the fie prit of the test oly allows terms that are 0). ANOTHER NOTE: You caot apply the alteratig series test because the sigs do ot alterate every term. Istead, ote that 0 si! 0 si! so sice 0 apple si apple, the th term of the series above is betwee 0 ad!. Sice P! coverges (by the ratio test, for example), our series coverges as well, ad does so absolutely. ( ) l This series coverges by the alteratig series test: The series is alteratig ad decreases ad teds to zero as l!. The series i fact coverges coditioally: Sice l diverges, P diverges as well. l k. X 3p + p, p l, so sice P p Here the divergece test works, sice the terms are /3, which teds to ifiity as!. The series diverges. l. X si( + ) + Here you eed to figure out what the sie term is doig, sice its iput cotais : Writig 3

4 m. out the first few terms: X si( + ) + si( ) (+) + si( 3 ) + ( ) si( 5 ) 5 + (+) si( 7 ) 0 + ( ) so the series is alteratig! Which is ot obvious from how it s writte. The absolute values of the terms are which is decreasig ad teds to zero as!, so the series coverges + by the alteratig series test. However! It also coverges absolutely: The absolute value series is just P +, which coverges by limit compariso with a p-series, for example. ( ) + Sice + is always odd, ( ) + is always equal to. This series is just is times the harmoic series, so it diverges. P, which. 0! ()! Agai, there are factorial ad expoetial terms, so this shouts do the ratio test! Set up: ( +)! + ()!! ( +)!! NOTE: (( +))!( +)!NOT ( +)!. Use safety paretheses whe chagig to ( +)i order to avoid mistakes. Next, we cacel (like terms): Expoetial cacelig is pretty simple: +. Factorial cacelig is a bit trickier, ad it helps to write them out if you re havig trouble: ( +)!! 3 ( +) 3 so everythig cacels except the ( +)upstairs: (+)!! ( +). For ()! (+)!, ()! ( +)! 3 () 3 ()( +)( +) so cacelig leaves us with (+)(+). 4

5 After all the cacelatio it remais to evaluate! ( +) ( +)( +) which is equal to zero, because this is a ratioal fuctio whose deomiator has a larger degree tha its umerator. Sice L 0, the series coverges absolutely by the ratio test. o. cos( ) Agai, the cosie cotais i its iput, so it behooves us to figure out what it is doig: cos( ) cos( ) + cos(4 ) + cos(6 ) + 3 (+) + (+) (+) 3 + So this is actually just the harmoic series P, which we kow diverges (by p-test or itegral test, for example). p. (arcta(/)) q. We are give the hit that arcta( ). Let s verify this asymptotic relatio! x x apple " arcta(/x) lim lim ( # apple apple (/x) + /x ) x lim x! /x x! /x x! (/x) + lim x! +x so they are asymptotic (above: first equality is L Hôpital s rule sice the origial limit is 0/0, secod equality is cacelig the ( /x )s, third equality is multiplyig by x /x ). After we have the hit, the series is easy to figure out usig the limit compariso test: The terms are asymptotic to (/) ad P coverges (by p-series test) ad so our series coverges as well (it does so absolutely, sice all its terms are positive). (l ) Like c., this oe requires the itegral test (oe of the two strategies we ve metioed that you should try whe a logarithm is i play): Usig the substitutio u lxad du /x, Z x(l x) du u u + C l x + C So, Z apple x(l x) lim b! l b l apple lim b! l Sice the itegral above coverges, our series coverges (absolutely). l b l 5

6 r. e The ratio test, sice there is a expoetial term this yields L /e, so the series coverges absolutely. s. ()!(3)! (5)! The ratio test, sice the series has factorials i it. Set up: Cacelig like factorials as before, ( +)!(3 +3)! (5)!! (5 +5)! ()!(3)! so, ( +)! ( +)( +) ()! (3 +3)! (3 +)(3 +)(3 +3) (3)! (5)! (5 +5)! (5 +)(5 +)(5 +3)(5 +4)(5 +5)! ( +)( +)(3 +)(3 +)(3 +3) (5 +)(5 +)(5 +3)(5 +4)(5 +5) How to take this limit? Should we expad everythig? No! Sice the umerator ad deomiator both have the same degree (which is 5), we oly eed to figure out what the leadig terms will be after we expad:! () (3) 3 +( ) (5) 5 +( ) lim! ( ) ( ) t. where the ( ) represet terms with lower powers of these wo t matter i the limit. We get L 08, which is <, so the series coverges absolutely si p Sice is ot iside the sie, this probably requires us to use the direct compariso test (or the absolute covergece test) with apple si apple. The umerator is at most 3 ad at least, so p apple +si p apple p 3. Sice P p diverges (by p-series test), our series diverges as well. 6