Sequeces ad Series Matt Rosezweig Cotets Sequeces ad Series. Sequeces.................................................. Series....................................................3 Rudi Chapter 3 Exercises........................................ 3.3. Exercise............................................. 3.3. Exercise............................................. 3.3.3 Exercise 3............................................. 4.3.4 Exercise 4............................................. 4.3.5 Exercise 6............................................. 4.3.6 Exercise 7............................................. 5.3.7 Exercise 8............................................. 5.3.8 Exercise 9............................................. 5.3.9 Exercise 0............................................ 6.3.0 Exercise............................................ 6.3. Exercise............................................ 7.3. Exercise 3............................................ 8.3.3 Exercise 4............................................ 9.3.4 Exercise 5............................................ 0.3.5 Exercise 6.............................................3.6 Exercise 7.............................................3.7 Exercise 8.............................................3.8 Exercise 9.............................................3.9 Exercise 0.............................................3.0 Exercise............................................ Sequeces ad Series. Sequeces Propositio.. For p > 0, lim p 0.. For p > 0, lim p. 3. lim. 4. For p > 0 ad α R, lim α (+p) 0. 5. For x <, lim x 0. Proof. I all the followig proofs, we assume that ɛ > 0 is give. For (), by the Archimedea property, we ca take ɛ p so that p (ɛ p ) p ɛ For (), fix p > 0. If p >, set x : p ad observe that x > 0. By the biomial theorem, p Hece, lim sup x lim first case by cosiderig p. For (3), set x : + x ( + x ) p x p 0. If p, the the result is trivial. If 0 < p <, the we retur to the. Observe that x 0 ad by the biomial theorem, ( + x ) ( ) x ( ) x x
By (), lim 0. For (4), fix α R ad p > 0, ad choose a positive iteger k > α. For > k, we have by the biomial theorem that ( ) ( + p) > p k! ( ) ( k + ) ( ) k p k ( k)!k! pk p k k > k! k! ( + p) < k k! k p k Hece, 0 < Sice k α > 0, k α 0 by (). For (5), take α 0 ad p x > 0 i (4).. Series α ( + p) < k k! k α p k The followig propositio attributed to Cauchy shows that the covergece of a series with mootoically decreasig terms is determied by the growth of a small subset of its terms. Propositio. (Cauchy s Covergece Test) Let ( ) be a decreasig sequece of real umbers bouded from below by 0. The the series coverges if ad oly if the series coverges. Proof. Suppose coverges, the k k a k ( a + k k ) a k a + ( k )a k, which implies the covergece of k k a k ad therefore k a k by the compariso test. If coverges, the k a k a + k+ j k + which implies that coverges by the compariso test. a j k a j, j k a k Cauchy s covergece test allows us to give a short proof of the covergece (ad divergece) coditios for the harmoic p-series. p Propositio 3. (Root Test) Let ( ) be a sequece of complex umbers, ad set α : lim sup. The. if α <, the series coverges;. if α >, the series diverges; 3. if α, the root test is icoclusive. Proof. Suppose α <. Sice fiitely may terms do ot affect the covergece of, we may assume that sup <. We see that is absolutely coverget by the compariso test, sice ( ) < α α α Aalogously, if α >, the there exists a subsequece of idices ( k ) k such that k coditio lim does ot hold. >, so the ecessary
To see that the case α provides us isufficiet iformatio to determie covergece or divergece, cosider the followig examples. If for all Z, the clearly diverges to, but for all. If, the lim sup ( ) lim, where the peultimate iequality follows from cotiuity, ad π 6. Propositio 4. (Ratio Test) For a sequece ( ) of complex umbers, the series a. coverges, if lim sup + < ;. diverges if a+ for all but fiitely may..3 Rudi Chapter 3 Exercises.3. Exercise We first prove the useful reverse triagle iequality. Lemma 5. For a, b C, a b a + b. Proof. Observe that a + b (a + b)(a + b) a + (ab + ba) + b a + Re(ab) + b Takig the square root of both sides completes the proof. a a b + b a b Let (s ) be a coverget sequece of complex umbers. For ɛ > 0, there exists N Z such that N implies s s. By the reverse triagle iequality, N s s s s < ɛ The coverse is false. Defie s ad s +. The (s ) oscillates betwee ad, but s for all..3. Exercise lim +. Proof. We ca write + + x, for x 0, so that + ( + x) + x + x x + x 0 x For ay 0 < c <, I claim that x c for all but fiitely may. Ideed, otherwise there exists a subsequee k such that k x + x k (c ) + c c + c < k 4 k 4 k for all k sufficietly large, which is a cotradictio. We see that Lettig c completes the proof. ( + ) + ( + x) + x ( c) 3
.3.3 Exercise 3 Set s :, ad for Z, defie s + : + s The the sequece (s ) coverges ad moreover, s < for all. Proof. By iductio, we see that s > > for all. Hece, s + + s s + < s + < I claim that s < s +. The base case follows from the mootoicity of the square root fuctio. Suppose the assertio holds for all j. The s + + s > + s s.3.4 Exercise 4 Defie a real sequece (s ) by The lim if s 0 ad lim sup s. Proof..3.5 Exercise 6 If : +, the for ay N Z, 0 s s : m m, m Z + s m m +, m Z so diverges to. N +, If : +, the usig our above estimate for +, we see that + < Hece, coverges by the compariso test. 3 Suppose : +z. I claim that the series coverges for z > ad diverges for z <. First, suppose z >. For ay N Z, + z z ( z ) k z z k z ( z ) z z which coverges as N by compariso with the geometric series. Now suppose that z <. 4
.3.6 Exercise 7 Suppose ( ) is a sequece of oegative real umbers such that coverges. The the series a coverges. Proof..3.7 Exercise 8 If coverges ad (b ) is a bouded, mootoically icreasig sequece, the b also coverges. Proof. By cosiderig real ad imagiary parts separately, it suffices to cosider the case where the are real. Without loss of geerality, we may assume that b > 0 for all. Set b : sup b. Let ɛ > 0 be give, ad choose N 0 Z such that N, M N 0 implies that N < ɛ. M+ sg( ) b M+ M+ b b ( N M+ ) < bɛ If 0 the b b sg( ) < b, ad if < 0, the sice b b, b b b b. Hece, ( N ) bɛ < b b We coclude that.3.8 Exercise 9 M+ M+ N M+ b < ɛ, so the series b coverges by the Cauchy criterio. The radius of covergece of 0 3 z is. Proof. Sice the fiite limit of products of sequeces is the product of the limits, we have that ( ) 3 lim sup( 3 ) lim sup ( ) 3 lim The radius of covergece of 0! z is ifiite. Proof. Applyig the biomial formula to, for large, we obtai the upper boud!! m! k!( k)! k!( k)! + k!( k)! m m e m + km+ k! + km+ km+ Sice e k k!, we ca choose N Z such that m, N implies that km+ k! < ɛ. Choose N > N such that N e implies that N < ɛ. The N! e N + kn+ k! k! k! < ɛ + ɛ ɛ We coclude that lim sup! 0, ad therefore the radius of covergece is ifiite. 5
The radius of covergece of 0 z is. Proof. Sice lim a 0 implies that lim a lim sup ( a, we have that ) lim ( ) (lim ) The radius of covergece 0 3 3 z is 3. Proof. By the same argumets used above, we have that lim sup ( 3 3 ) ( ) 3 lim (lim ) 3 3 3 3.3.9 Exercise 0 Let ( ) be a sequece of itegers, ifiitely may of which are ozero. The the radius of covergece of z is at most. Proof. Deote the radius of covergece of the series z by R [0, ]. If z coverges for some z C, the lim z 0 If z >, the for all sufficietly large, <. Sice Z, 0, for all but fiitely may idices, which cotradicts our hypothesis..3.0 Exercise Let ( ) be a sequece of positive real umbers. Suppose diverges. The + diverges. Proof. If + coverges, the a + 0, so for ɛ > 0 small, + < ɛ < ɛ + ɛ < ɛ ɛ If ( ) is ubouded, the ( ) is bouded by some positive costat M >. The a + does ot ted to 0 as, hece Hece, + diverges by the compariso test. If s : a + +, the > a ( + a ) + + a N ( + a N ) > + M + M + + does ot coverge. Suppose a N+ + + a N+k s N, s N+ s N+k s N+k k Z ad therefore s diverges. Proof. For ay k Z, s N+ < < s N+k. Hece, a N+ s N+ + + a N+k s N+k a N+ + + a N+k s N+k This lower boud shows that the series s s N+k s N s N+k s N s N+k diverges by the Cauchy criterio. 6
The sequece ( a s ) satisfies ad therefore the series s Proof. Sice s < s, we have that coverges. s s s s < s s s s s s s s Sice s ad the RHS i the above iequality is telescopig, we see that as N. The series ( ). s a s + [ ] a ( s s s + ) a s s N s + s + coverges. The covergece or divergece of Proof. The secod claim follows from otig that coverges by the compariso test. For the first claim, first cosider :. The + depeds o the sequece + ad that <. Hece, +, which diverges sice the harmoic series diverges. Now defie by { : m, m Z otherwise + I claim that diverges. Let α > 0. Sice diverges, we ca choose a iteger N sufficietly large so that M N implies that M > α. The But + coverges sice > N m m > α + N m,m Z + N m,m Z + m m + < May thaks to the math.stackexchage commuity for helpig me out with this last part..3. Exercise Suppose ( ) is a sequece of positive reals such that the series coverges. Defie tails If m <, the r : a m, m Z ad therefore r coverges. a m r m + + r > r r m 7
Proof. First, ote that r + < r ad sice coverges, r 0. Hece, a m + + > a m + + r m r + r + > r r m r r m r m r m r m r m where m <. Let ɛ > 0 ad m Z be give. Sice r 0, we ca choose > m sufficietly large so that r r m > ɛ, which implies that the series r does ot satisfy the Cauchy criterio. For Z, so that the series r coverges. r < ( r r + ), Proof. We ca write r r +, so that r ( r + r + ) ( r + r ) r + r r r sice r+ r <. For ay N Z, as N. Hece,.3. Exercise 3 r r < coverges. ( ) r+ ( + r ) r + r < ( r r + ), ( r r + ) ( r r N+ ) 0 If a k ad b k coverge absolutely, the the Cauchy product c k coverges absolutely ad ( ) ( ) c k a k b k Proof. Let ɛ (0, ) be give. Set A a k ad B b k. Choose M > 0 such that a k, b k M, choose m > 0 such that max {sup k a k, sup k b k } m, ad choose N 0 N such that { } N N 0 max a k, b k < ɛ kn+ kn+ The for N N 0, we have N AB ( ) ) ( ) N ( ) N c k A a k (B b k + A a k b k + B b k a k + ( < Mɛ + Mɛ + a i ) b j < 4Mɛ in jn To see absolute covergece, observe that for N N, ( k N c k a i b k i a i ) ( b j a i ) b j < i0 i j i j N<i+j N Sice the partial sums N c k are mootoically odecreasig, they coverge by the mootoe covergece theorem. It follows by iductio that the Cauchy product of a fiitely may absolutely coverget series a k,j, j r, is absolutely coverget with limit a k,a k, a kr,r 0 k + +k r a i b j 8
.3.3 Exercise 4 Let (s ) 0 be a sequece of complex umbers. For Z 0, we defie the th arithmetric mea σ by If lim s s, the lim σ s. σ : s 0 + s + + s + Proof. Choose ɛ > 0, ad let N Z be sufficietly large such that N implies that s s < ɛ. The σ s s 0 + s + + s s + s 0 + s + + s s ( + ) + + s 0 s + + s s Choose N N such that N + N j0 s j s < ɛ. The N σ s j0 s j s + + jn+ s j s + N j0 s j s N + + ɛ < ɛ Note that a sequece (s ) eed ot coverge i order for its arithmetic meas (σ ) to coverge. Let s : ( ). The { 0 (mod ) σ + 0 (mod ) from which it is immediate that σ 0. We ca eve costruct a sequece (s ) satisfyig s > 0 for all, lim sup s, yet lim σ 0. Defie { k k, for some k Z 0 s : k ( k, k+ ) For a give Z 0, let k k() be the maximal iteger such that k, so that σ s k 0 + s + + s j0 j + j j k(k + ) + 4( (k+) ) + + ( + ) as. For, set : s s. The Proof. Observe that s σ s 0 + [s k s k ] + k s σ + s k s 0 + k k + k ka k k a k s 0 + + a k + a k + ka k k k k j a j ( k)a k k k(k + ) + 4 ( k + ) s 0 + k 0 k [s j s j ] j 9
Now assume that lim 0 ad lim σ σ exists. The lim s σ. Proof. Let ɛ > 0 be give. Choose N Z sufficietly large so that N implies that < ɛ ad σ σ < ɛ. Choose N N such that N implies that The + k a k < ɛ k N s σ s σ + σ σ < ɛ + + k ka k < ɛ + + < ɛ + + 3ɛ kn+ kn+ k a k ɛ It turs out that we ca relax our hypothesis above that lim 0 to just ( ) is a bouded sequece. Proof. Choose M > 0 such that < M for all. Let Z be large. For m <, write s σ m m s σ σ + m im+ ( m) i0 s i ( + )( m) s i + m im+ [s s i ] + ( + ) im+ s i + ( m)( + ) m im+ (m + )s i ( m) m i0 s i + ( m)( + ) m m + m σ m + m σ m + m im+ [s s i ] im+ im+ [s s i ] [s s i ] For i m +, s s i ki+ Choose ɛ > 0 small. The ɛ +ɛ a k ki+ a k ki+ M k ki+ M M( i) M( m ) i + i + m + lies i the iterval [m, m + ), for some oegative iteger m, so that m( + ɛ) ɛ < (m + )( + ɛ) (m + )ɛ m < + (m + )ɛ, which implies that m+ m ɛ ad therefore s s i < Mɛ, for m + i. Sice m as ad by our hypothesis that σ σ, we see that lim sup s σ lim sup Sice ɛ > 0 was arbitrary, we coclude that s σ..3.4 Exercise 5 [ σ σ m + m Mɛ + lim sup σ σ m Mɛ im+ For solutios to exercises 6, 7, ad 8, see my ote o square root algorithms. Mɛ ] 0
.3.5 Exercise 6.3.6 Exercise 7.3.7 Exercise 8.3.8 Exercise 9.3.9 Exercise 0 If (p ) is a Cauchy sequece i a metric space (X, d), ad some subsequece (p k ) k p X. The p p. coverges to a poit Proof. Choose N Z such that, m N implies that d(p, p m ) < ɛ. Choose k 0 Z such that k k 0 implies that k N ad d(p k, p) < ɛ. By the triagle iequality, d(p, p) d(p, p k ) + d(p k, p) ɛ + ɛ ɛ.3.0 Exercise For solutios to exercises 3, 4, ad 5, see my blog post etitled How to Complete Your Metric Space.