Complete Solutions to Supplementary Exercises on Infinite Series

Transcription

1 Coplete Solutios to Suppleetary Eercises o Ifiite Series. (a) We eed to fid the su ito partial fractios gives By the cover up rule we have Therefore Let S S A / ad A B B. Covertig the suad / the by usig these partial fractios we have: The ifiite series is give by li S li li P a g e 6

2 The give series coverges with the su. (b) We are give 6. Let S S The ifiite series is be the partial su, the li S li (*) Note that each of these are geoetric series with coo ratio ad : / li ad / Substitutig these ito the above (*) gives (c) We eed to fid fractios gives / li / li S li Multiplyig through by Substitutig Substitutig 0. Covertig the suad ito partial A B C D yields A B C D ( ) ito ( ) yields D 0 0 D ito ( ) gives Equatig coefficiets of Equatig coefficiets of 0 A 0 B 0 B i ( ) gives 0 A C A C i ( ): P a g e 6

3 0 A B C D Substitutig the above evaluated B, D ito this gives 0 A C 0 A C Fro the two siultaeous equatios ivolvig A ad C: A C ad 0 A C we have A 0, C 0 Therefore our partial fractios coversio is The th partial su is give by S The su of the ifiite series is give by li S li Therefore. P a g e 6

4 (d) We eed to fid partial fractios gives. Covertig the suad ito A B C D Multiplyig this by gives A B C D (*) Substitutig Substitutig ito (*) gives A0 B C 0 D 0 4B B 8 ito (*) gives A0 B 0 C 0 D 4D D 8 Substitutig 0 Puttig B 8 ad Equatig coefficiets of ito (*) gives D 8 0 A B C D ito this yields 0 A C A C 8 8 i (*): 0 8A 8C A C 0 Fro the last two equatios we have A 0 ad C 0. Substitutig these evaluated A, B, C ad D gives the partial decopositio: The th partial su S Epadig this out S is equal to 8 P a g e 4 6

5 S The ifiite su is give by li S li 8 8 (e) We are asked to fid ta. We first fid the partial su ta I order to evaluate this partial su we fid the su for : Now we fid the su for : ta ta ta 8 ta 8 By hit 8 0 ta ta 5 P a g e 5 6

6 ta ta ta ta ta 8 Siilarly we fid Ca you spot a patter? by previous result 8 9 ta ta ta 4 ta 5 ta 4 ta ad 5 ta You ca prove this by iductio. The ifiite su is give by 5 ta 6 ta li ta li ta li ta ta / 4. (a) We eed to use the copariso test to fid whether coverges or ot. Cosider the suad: Now Because for is a geoetric series with the coo ratio r ad sice r so coverges. P a g e 6 6

7 Hece by the copariso test (b) We eed to test give by For 0 we have si Eaiig the ifiite series This coverges. si for covergece. The power series for si is 5 si! 5!. Applyig this to si gives si gives of course is a geoetric series with coo ratio Sice the coo ratio is less tha so coverges. Hece by the copariso test we coclude that the give (c) We are asked to test fractios: Cover up gives gives A r. si coverges.. Covertig the suad ito partial A B ad B. Substitutig these ito the above Fro this we have. Now P a g e 7 6

8 By the copariso test we have (d) How do we test diverges ta 4 for covergece? diverges. We eed to fid a iequality ivolvig the taget fuctio. The power series epasio of ta is give i forula (7.8): (7.8) ta Fro this we have The ifiite series provided 5 5 ta 4 ta 4 diverges. (e) We eed to test suad is By the p test we have coverges. (f) We are give Also we kow that diverges. for 0. For the give suad we have ta 4 4 diverges so by the copariso test we coclude that for covergece. A iequality ivolvig the Because coverges so by the copariso test. We have the iequality (g) We are asked to test l gives: diverges so by the copariso test l. Coparig the graphs of ad P a g e 8 6

9 Fro this we ca see that l idetity.) Fro this l we have The haroic series l diverges. (h) How do we test for. (It is a well kow l l so diverges so by the copariso test the give series for covergece? We first covert the suad ito partial fractios: A B C Fatorisig deoiator Multiplyig both sides by gives A B C Substitutig (*) ito (*) yields A B C 0 A A Substitutig 0 ito (*) gives ( ) P a g e 9 6

10 Equatig coefficiets of A C C C i (*) gives A B B B Substitutig A, B ad C ito ( ) gives The ifiite series diverges so diverges ad so by the copariso test (i) We are asked to test ter that is beig squared: Squarig this gives Now each of these diverges. for covergece. Let us first cosider the Because 4 By the copariso test (j) We are give 6 6 4, ad coverge by the p test. So 6 4 coverges 6 4 coverges.. We have the iequality Applyig the iequality y y for 0, y 0 we have P a g e 0 6

11 Takig the reciprocal gives The ifiite series give series (k) We are asked to test diverges so by the copariso test we coclude the diverges. l 4 5 Fro the well-kow iequality: We have l for covergece. How? for 0 l /5 /5 Applyig the followig logarith law l l l l 5 5 /5 /5 Usig this o the uerator of the suad gives Now by the p test, series l 4 5 /5 l coverges. 5 /0 (l) We are asked to test 5/ to the left had side: coverges so by the copariso test the give. Cosider the suad Multiplyig by For the atural uber we have Takig the reciprocal we have the iequality P a g e 6

12 By the p test diverges so copariso test we coclude that () We are asked to test diverges. Therefore by the diverges.. Cosider the suad: Cosider the bracket ter of the deoiator of the last epressio: for Substitutig this iequality ito the above gives By the p test, / / coverges so copariso test we coclude / coverges. Hece by the coverges. /. We apply the ratio test i each case for this questio. (a) We are asked to test! for covergece. Let a! the a. Usig the ratio test! a li a gives P a g e 6

13 L li!!! li!! li li 0! Therefore the give series (b) Siilarly for Fidig the liit; By the ratio test! we let a the a coverges li L li li because coverges because L. (c) We are asked to test ta for covergece. Let ta a so a ta ad ta ta L li li ta ta ta li li ta Eaiig the fractio o the right had side: * P a g e 6

14 (d) ta ta ta where ta ta ta Usig the double agle forula of chapter 4: (4.55) taa O ta gives ta Substitutig back ta ta A A ta / ta ta ta / ta / ta / ta / ta / ta / Evaluatig the liit yields gives ta ta ta li ta 0 Puttig this ito (*) gives ta L li li ta Sice L so by the ratio test ta coverges. 5 We eed to test. Let 54 P a g e 4 6

15 (e) (f) a Evaluatig the liit the a L li a a 5 5 li li li li Sice L We have to test so the give series Evaluatig the liit Sice for covergece. Let a the a li L li li li L so the give series We are give the series a! Workig out the liit gives coverges.. Let! the a coverges.! P a g e 5 6

16 (g) (h) L!! a!! li li li li li a Sice L so the give series We are asked to test Fidig the liit a!. Let! the a!! coverges.!!!! Because! L li li! li li 0 Dividig uerator ad deoiator by Sice L 0 we coclude by the ratio test, We are asked to test a!. Let!!! Evaluatig the liitig ratio the a!!! coverges. P a g e 6 6

17 (i)!! li L!! Because li!!!!!!!! li Hece by the ratio test the give series Agai usig the ratio test to see whether Let a si the a! coverges.! si coverges or ot. si. We have si L li si si si li li li where si si Now usig the trigooetric idetity (4.5) sia sia cosa Fro this we have si si cos. Substitutig this ito the above o the etree right had side ter gives si si si si cos cos Substitutig back ad evaluatig the liit o the right i the above P a g e 7 6

18 si li li si cos li cos Because li 0 so l i cos cos0 si Substitutig this li ito the above gives si si L li li si By the ratio test we coclude that si coverges. 4. I this questio we use the itegral test for covergece or o covergece. Sice we will be dealig with a defiite itegral (iproper itegral) so we will igore the costat of itegratio i our workig. (a) The for We are give the series we have f l. Let l l (i) f 0 (Positive) because ad l (ii) f is decreasig because l (iii) f is cotiuous because ad l therefore l is cotiuous which iplies f The graph of this fuctio is give by: are both positive. is icreasig. are both cotiuous is cotiuous. P a g e 8 6

19 We ca ow apply the itegral test because all three coditios are satisfied. Cosider the iproper itegral d d li M l l How do we itegrate this fuctio? By substitutio. Let u l the du d du d First fidig the idefiite itegral we have d l Workig out the iproper itegral du u M u du u C C l M li d li M l l li M l l l M M M P a g e 9 6

20 Sice the iproper itegral coverges so by the itegral test the give series l coverges. (b) Now we are asked to use the itegral test o the series Let f the for we have l 0 ad l 0 for. (i) f 0 (Positive) because (ii) I order to use the itegral test we eed to check that fid the derivative of For f :. l f 0 l l f l l l we have l 0 ad l Therefore (iii) f f >0 so l f 0 l is a strictly decreasig fuctio. f is cotiuous because ad l is cotiuous. is decreasig. We l are cotiuous for ad so As all three coditios are satisfied so we ca apply the itegral test. Cosider the iproper itegral: d li d l l M M Agai we use itegratio by substitutio with Chagig the liits; du d du d Whe M the u lm ad whe We have u l the l. Differetiatig this gives u. P a g e 0 6

21 lm d li du l M u lm du lm li li lu M u M l l li llm ll M l This iproper itegral diverges so by the itegral test we coclude that the give series l diverges. (c) We are asked to test the series For (i) f 0 we have f. Let [Positive] because we have a square fuctio. (ii) Fidig the derivative of Now for fro f we have f the uerator is positive. All the other ters apart i the above are positive so f 0 which iplies that f strictly decreasig fuctio. (iii) Sice ad are cotiuous fuctios so is cotiuous iplies f is cotiuous All three coditios of the itegral test are satisfied. Cosider the iproper itegral is a P a g e 6

22 M d li d M Let us first eaie the idefiite itegral: d Covertig the itegrad ito partial fractios gives A B C D Multiplyig through by Equatig coefficiets of : 0 A : B gives A B C D : A C 0 C C d ( ) (*) Cost: B D D D 0 Substitutig these values A 0, B, C ad D 0 ito ( ): Itegratig this yields d d d du ta u stadard itegral (8.6) substitutig u ta u ta Now evaluatig the iproper itegral fro above P a g e 6

23 M M li d li ta M M li ta M ta M M The iproper itegral coverges therefore by the itegral test we coclude that the give series (d) Let coverges. We eed to test l for covergece. f l. For we have (i) f 0 [Positive] because for the atural log of this is positive ad of course the logarithic arguet is positive. so (ii) To fid the derivative of f Applyig the quotiet rule we have For f Multiplyig uerator ad deoiator by we have f 0 the fractio is positive. Hece f we rewrite f l l l by usig the laws of logs as l l 4 l l because of egative sig i frot of the fractio ad f is strictly decreasig. (iii) We have is cotiuous ad l is cotiuous therefore f l is cotiuous. P a g e 6

24 We ca ow apply the itegral test. We eed to test whether eaie the idefiite itegral: l d l d coverges or ot. Let us first How do we itegrate this fuctio? By parts. Let u l ad / v. Differetiatig oe ad itegratig the other gives u By (8.0) l i reverse u / / v d Applyig the itegratio by parts forula gives l d uv u vd l 4 d l 4 d(*) We eed to fid the itegral o the right had side of (*). How? Use itegratio by substitutio with p. The dp d dp pdp d Therefore p d p dp p p d p 4 * * p 4 Covertig the itegrad i (**) to partial fractios we have p p Ap B Cp D 4 p p p p p Multiplyig both sides by p p gives ( ) P a g e 4 6

25 Substitutig p Equatig coefficiets of p Ap B p Cp D p ito this gives p : 0 A C A C p : B D p: 0 A C A C Fro the Substitutig A 0 Substitutig A B A B p ad p coefficiets we have A C 0 B Puttig these A C 0, ito first equatio gives. B B ito the above coefficiets of p D D B p p p Substitutig this result ito (**) gives 4 p gives ad D ito ( ): p p p d dp dp dp ta p p p p The itegral o the right had side ca be writte as by (8.6) p dp dp By partial fractios p p p p Puttig this ito the above l p l p P a g e 5 6

26 d lp lp ta p l l ta Because p l ta Puttig this d l ta ito (*) gives l d l 4 d l l 4 ta The iproper itegral is give by l d li l l 4 ta M M M li M l l 4 ta M M M M l l 4 ta M Now li l li l M l 0 M M ad siilarly M M M li l 0 M M 4 li ta 4. Substitutig these ito the above gives M l d l l 4 ta Hece the iproper itegral coverges so the give series l Also M coverges. M P a g e 6 6