Limit superior and limit inferior c Prof. Philip Pennance 1 -Draft: April 17, 2017

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1 Limit erior ad limit iferior c Prof. Philip Peace -Draft: April 7, 207. Defiitio. The limit erior of a sequece a is the exteded real umber defied by lim a = lim a k k Similarly, the limit iferior of a is defied by lim if a = lim if a k k 2. Existece: Let T i = {a k : k i} be the i th tail. Notice that T T 2 Hece T m is always o-icreasig ad if T m o-decreasig. It follows that ad lim a = lim T = if T lim if a = lim if T = if T always exist as exteded real umbers ad that lim if a lim a 3. Remark: For all m, IN: 4. Notatio. lim a ad lim if a are ofte deoted lim ad lim respectively. 5. Example. Let a = if is odd if is eve The T = / ad if T =. Hece lim a = 0 lim if a =. 6. Claim. Let a be a bouded sequece ad A, A IR such that the A < lim a < A a a > A for at most fiitely may. b a > A for ifiitely may. if T if T max{m,} T max{m,} T m which also implies equatio. A lim A Exercise a for fiitely may a for ifiitely may

2 7. Alterative characterizatio of the limit erior. Let a be a sequece. The lim a is the uique exteded real umber umber L with the property that for all A, A IR with A < L < A, the followig hold: a a > A for at most fiitely may. b a > A for ifiitely may. The previous claim shows that L = lim a satisfies the property. If the property is also true for L L it is a easy exercise to se that a obvious cotradictio arises: L L a for fiitely may a for ifiitely may 8. Corollary. The limit erior of a is the smallest real umber L such that, for ay positive real umber ɛ there exists a atural umber N such that a < L + ɛ for all > N Exercise. 9. Dual property. lim if a is the uique exteded real umber L such that for all A, A IR with A < L < A, the followig hold: a a < A for at most fiitely may. b a < A for ifiitely may. 0. Corollary. The limit iferior of a is the largest exteded real umber L such that, for ay positive real umber ɛ there exists a atural umber N such that a > L ɛ for all > N.. Defiitio. Let α = a be a bouded sequece. A elemet b IR is a evetual upper boud of a α if b < a for at most fiitely may. 2. Let B α be the set of all evetual upper bouds of α. The a B α cotais all upper bouds of α UBα B α b B α is absorbig from the right, i.e., x > b B α x B α c if B α = lim a Proof of c. Let B = if B α ad pose that A, A IR satisfy A < B < A The there exists b B α such that B b < A. It follows by right absorptio that A B α ad so A < a for at most fiitely may. O the other had A < if B α A B α A < a for ifiitely may Thus B satisfies the characterizatio of the limit erior i 7. Thus the limit erior of a sequece is the ifimum of the set of evetual upper bouds. 3. Warig. Some sources describe the limit erior as a evetual upperboud This termiology is misleadig. For example, lim = 0 yet 0 is a lower boud for the sequece. 4. Exercise. Use the alterative characterizatios of limit erior ad limit iferior to give aother proof of. 5. If lim a exists the lim if a = lim a = lim a Notice the lim a satisfies the alterative characterizatios of the limits erior ad iferior give i 7 ad 8.

3 6. If lim if a = lim a the the sequece a coverges to their commo value. From 7, 8 above, it follows that ay ope iterval cotaiig the commo value cotais all but a fiite umber of terms of the sequece. 7. Claim. lim a +b lim a +lim b 2 Proof: Recall that for ay two sequeces a ad b {a + b } a + b. I particular, applyig this to the -th tails of the two sequeces k {a k + b k } k α + β. k Sice this holds for all the limit ca be take o both sides: lim a + b = lim lim a k + b k k a k + k k b k = lim a + lim b Alterative Let α = lim a ad β = lim b. The N, N such that ad so > N a < α + ɛ/2 > N b < β + ɛ/2 > N + N a + b < α + β + ɛ α + β + ɛ a + b for fiitely may It must be that lim a + b α + β + ɛ. Sice ɛ was arbitrary lim a + b α + β ad the result follows. 8. Mea Property Let a be a bouded sequece. Defie x by The x = a + a a lim x m lim a with a obvious dual statemet for the limit iferior. Let ɛ > 0. The, there exists N such that > N x < lim a + ɛ Let U = lim a ad > N.The x < a + a a N + N[U + ɛ] 3 Notice that for all sufficietly large, x < U + ɛ This implies that lim x U. Remark: This also follows by applyig property 2 to equatio Bolzao Weierstrass Theorem. A bouded coverget sequece has a coverget subsequece. Let a be bouded. The idea is to costruct a subsequece covergig to L = lim a. such that a > L. Proceedig by recursio, pose that there exist atural umbers < 2 < < k

4 such that a j > L /j, j k It is ot assumed that the a j are icreasig. However, sice a k < L /j for fiitely may k a k a k for ifiitely may k for fiitely may k L j it follows that L lim if a k L ad so L lim if a k lim a k L provig that the subsequece coverges to L. 20. Corollary. Let a be bouded. The lim a is the maximum of the set of subsequetial limits. The existece of a larger subsequetial limit would cotradict the evetual upper boud property. 2. Claim. [Sequetial Completeess] Every Cauchy sequece of real umbers coverges. Let a be Cauchy. Let ɛ > 0. There exists N such that a a m < ɛ for all, m N. I particular, a a N < ɛ for all N. a for fiitely may a N ɛ a N + ɛ a for fiitely may It follows that a is bouded ad moreover: a N ɛ lim if a lim a a N + ɛ Therefore lim a lim if a < 2ɛ. Sice ɛ was arbitrary the limits must be equal provig that a coverges. 22. Claim [Cauchy Hadamard] A real or complex power series a x has radius of covergece give by If R =. lim a lim a / {0, } ad x < R, the x lim a < There exists t < with x lim a < t It follows that for all sufficietly large x a t Hece a x t. Sice t < series coverges by compariso with the geometric series. O the other had, if x > R, the It follows that the x lim a > x a > for ifiitely may terms. Sice the th term does ot ted to 0 the series does ot coverge. Fially, if the sequece a / is ubouded the power series coverges oly at 0, while if the lim is 0 the series coverges o the etire plae ad R =. 23. Claim. Let a be a sequece of positive umbers, the a+ lim if lim if a a lim a lim a+ a By duality it suffices to prove the fial iequality. Let L = lim a+ a. Let A > L. The, there exists N such that

5 a + a < A for all > N. It follows by iductio that a N+k < A k a N for all k IN. ad so an a < A, A N > N Hece a < an A N / A, > N Takig the limit erior of both sides ad recallig that lim p = for all p > 0 gives lim a A. a+ Sice A > lim was arbitrary, it follows that a lim a lim a+ a 24. Corollary. Let a be a positive sequece. If lim a + = l where l = a is ot excluded the lim a = l. 25. Refiemet of d Alembert s ratio test. Cosider, oce more, the real or complex power series a x. The ratio test states that if a is a sequece with a 0 for all sufficietly large ad lim a + a = L the a if L <, the series coverges absolutely; b if L >, the series diverges; If L = the the test is icoclusive. If L does ot exist, the test ca be refied by otig that a if lim a + a <, the series coverges absolutely; b if lim if a + a >, the series diverges. 26. Remark: If the limit L = lim a + a exists the the radius of covergece of power series a x is also give by R = /L. If L does ot exist it might happe that L < lim a = R i which case, the ratio test uderestimates the radius of covergece. For example, if { 2 if eve, a = 2 2 if is odd. the exercise a x covergece R = yet 2 lim a + a = 4 has radius of which merely guaratees covergece for x < /4