MATH 312 Midterm I(Spring 2015)

Transcription

1 MATH 3 Midterm I(Sprig 05) Istructor: Xiaowei Wag Feb 3rd, :30pm-3:50pm, 05 Problem (0 poits). Test for covergece: p, p 0. (coverges for p < ad diverges for p by ratio test.). ( coverges, sice (log log ) log > log for > (log ) log e 0, hece (log ) log >. Our claim follows from compariso test.) ( ). (coverges as alteratig series sice 0 as.) ( ) log log. (coverges as alteratig series log log 0 as.) Problem (0 poits). Let {a } be a real-valued sequece. State the defiitio of lim sup a. Prove that lim sup a = U if ad oly if the followig is true (a) For ay ɛ > 0, there are ifiite may a s such that a > U ɛ. (b) For ay ɛ > 0, there are oly fiite may a s such that a > U + ɛ. Proof. Recall the defiitio: U = lim sup a if (A) ɛ > 0 ad m > 0, > m s.t. a > U ɛ.

2 (B) ɛ > 0, N s.t. for > N implies a < U + ɛ. :First, we otice that (B) implies (b). To obtai (a), let us apply (A) to ɛ > 0 ad m = 0, the there is > 0 s.t. a > U ɛ. Now apply (A) to ɛ > 0 ad m =, the there is a > s.t. a > U ɛ. Cotiue this process we obtai ifiite may {a k } s satisfyig a k > U ɛ, which is exactly the statemet of (a). : Clearly, we otice that (b) implies (B). To obtai (A), for fixed m, sice we have ifiite may a s satisfyig a > U ɛ, there is a > m such that a > U ɛ, ad our proof is completed. Problem 3 (0 poits). Let {a } be a real-valued sequece ad let σ = (a + + a )/. Show that lim if a lim if σ lim sup Proof. By symmetry, we will oly prove lim sup σ lim sup a. σ lim sup a. If lim sup a = the by Theorem 8.3 (d) we kow lim a =. So for ay give M > 0, N > 0 s.t. for N, we have a M. This implies σ = (a + + a N ) + + a = a + + a N + N ( M) = a + + a N + a N+ + + a from which we deduce that lim sup σ M. Sice M is arbtrary, we deduce that lim sup σ. For the case that lim sup a = a <. Give ay ɛ > 0, N > 0 s.t. for N, we have a < a + ɛ. Let > N, we have σ = (a + + a N ) + + a = a + + a N + N (a + ɛ) = a + + a N + a N+ + + a which implies that lim sup σ a + ɛ ad hece lim sup σ a sice ɛ is arbitrary. Problem (0 poits). Show the followig real-valued sequece {a } R coverge.

3 Proof. Notice that a <, a + a + 6 a3 + a 3. a + a + 6 a3 + a 3 6 a + + a a + + a a + a 3 6 a + a a + a this implies that So a + a a a 3. a +k a k a +j a +j j= ) +j k ( j= ( ) 0, as. By Cauchy coditio of sequeces, we obtai that {a } coverges. Problem 5 (0 poits). Prove that there are costats A, B R such that the followig formulas hold... k= k= log = log log + A + O( log ) = + B + O( ) Solutio. Apply Theorem 8.3 to the followig fuctios: 3

4 . Let f(x) = x log x 0 as x.. Let f(x) = x 0 as x. Problem 6 (0 poits). Give a coverges absolutely. Show that each of the followig series also coverges absolutely.. a. Proof. Sice a coverges, by Cauchy coditio we have lim a = 0. I particular, this implies a M. This implies that a M 3 a bouded above sice a coverges.. a a + a Proof. Agai the covergece of a implies lim a = 0. I particular, there is a N > 0 such that a / < for N. This implies that for N we have a a + a a. 6 By compariso test, we kow a a + a coverges absolutely. Problem 7 (0 poits). Let {a = real umber x: si x } ad {b = cos x } with a fixed

5 . Prove k= k= to ad use it to deduce x si cos kx = cos( + ) x si x ad e ikx = si(x/) si(x/) ei(+)x/ si kx = k= si x si( + ) x si x. Proof. It is i the book page 95.. Prove both a ad b are coverget. Proof. For a fixed x 0, we have cos kx = si x cos( + ) x si x si x ad k= si kx = si x si( + ) x si x si x. k= The covergeece of a ad b the follows from the Dirichlet s text, sice 0 as. 3. [Bous] Prove a is diverget. (Hit: ) si x si x for all N. ) si cos x x = ) Proof. Suppose a coverget the by compariso test we obtai si x also coverges absolutely. But si x = si x a = cos x = si x cos x diverges by part ), a cotradictio. So our assumptio that a is absurd. 5