Practice Problems 12 : Comparison, Limit comparison and Cauchy condensation tests P1 1. Let an ≥ 0 for all n 2 N. If n=1 an converges then show that P1 2 (a) n=1 an converges. Is the converse true ? P1 p (b) n=1 anan+1 converges. p P1 an (c) n=1 n converges. n P1 an+4 (d) n converges using comparison or limit comparison test. n=1 an+5 P1 1 2. Let (an) be a sequence such that an > 0 for all n and an ! 1. Show that n n=1 an converges. P1 P1 P1 2 3. Let n=1 an be a convergent series. Show that n=1 janj diverges if n=1 an diverges. P1 a1+a2+ ::: +an 4. Let an > 0 for all n 2 N. Show that the series n=1 n diverges. P1 P1 5. Assume that n=1 an and n=1 bn are convergent and an; bn ≥ 0 for all n 2 N. Show P1 p 2 2 that n=1 an + bn converges. Does the converse hold ? P1 2 P1 2 P1 p 6. Let an; bn 2 R for all n and n=1 an and n=1 bn converge. Show that n=1(an − bn) converges for all p ≥ 2. P1 nπ 7. Show that n=1 sin n+1 diverges. 3 2 P1 pan 8. Let an ≥ 0 for all n and n an ! ` for some ` > 0. Show that n=1 n converges. 9. Suppose a > 0 for all n and P1 a converges. Show that the series P1 1 − sin an n n=1 n n=1 an converges. P1 1 1 10. Consider the series n=1 an where an = n for n = 1; 4; 9; 16; ::: and an = n2 otherwise P1 (i.e., if n is not a perfect square). Show that n=1 an converges but nan 9 0. P1 11. Let (an) be a sequence of positive real numbers such that an+1 ≤ an for all n and n=1 an P1 converge. Show that n=1 n(an − an+1) converges. P1 1 12. Show that n=4 n(ln n)(ln(ln n)) diverges. P1 13. In each of the following cases, discuss the convergence/divergence of the series n=2 an where an equals: 1 sin( ) 2 1 p n lnpn 1 −n (a) (ln n)p ; (p > 0) (b) n (c) n (d) n2−ln n (e) e −1 (f) 1 (g) tan 1 (h) 1 − cos π (i) (ln n) sin 1 (j) tanp n 1+ 1 n n 2 n n n n n p p 1 3+cos n 2+sin3(n+1) n+1− n (k) (n + 2)(1 − cos n )(`) en (m) 2n+n2 (n) n P1 14. (*) Suppose that an > 0 for all n and n=1 an diverges. Let (An) be the sequence of partial sums of P1 a and (S ) be the sequence of partial sums of P1 an n=1 n n n=1 An (a) Show that (Sn) does not satisfy the Cauchy criterion. (b) Show that there exists a sequence (bn) such that bn+1 ≤ bn for all n, bn ! 0 and P1 n=1 bnan also diverges. Practice Problems 12 : Hints/Solutions 2 − 2 1. (a) Since an ! 0, an ≤ an eventually. Converse is not true: Take an = n 3 . p 1 (b) Use the inequality anan+1 ≤ 2 (an + an+1). q 1 1 1 (c) Use an n2 ≤ 2 (an + n2 ). n n n an+4 an+4 1 n 4 n 4 n an+4 5 n (d) Use n ≤ n ≤ + or apply LCT with , i.e., find lim n . an+5 5 5 5 5 n!1 an+5 4 1 1 2. Observe that n < n eventually. an 2 2 3. Since an ! 0, an ≤ janj eventually. a1+a2+ ::: +an a1 4. Note that n ≥ n . 2 2 2 p 2 2 5. Use the inequality an + bn ≤ (an + bn) . Converse is true because an ≤ an + bn. P1 2 p 2 6. It is sufficient to show that n=1(an − bn) converges because jan − bnj ≤ (an − bn) P1 2 2 eventually for p > 2. For convergence of n=1(an − bn) , use the inequality (a − b) = 2a2 + 2b2 − (a + b)2 ≤ 2a2 + 2b2. 1 nπ 7. Use the LCT with n : n sin n+1 ! π. 2 3 p 1 pan n 2 8. Use the LCT with n2 : n 1 = ann ! ` > 0. 2 1 sin an an−sin an 1 9. Use the LCT with a : 2 1 − = 3 ! . n an an an 6 1 1 1 1 1 1 1 1 1 10. The series is 1 + 22 + 32 + 4 + 52 + 62 + 72 + 82 + 9 + ::: The sequence of partial sums is 1 1 1 1 1 P1 1 bounded above by ( 22 + 32 + 52 + :::) + (1 + 4 + 9 + :::) ≤ 2 n=1 n2 but nan = 1 when n is a perfect square. P1 11. The partial sum Sn of n=1 n(an − an+1) is a1 + a2 + ::: + an − nan+1. 12. Use the Cauchy condensation test and the fact that ln 2 < 1. 1 n 13. (a) Diverges (Use the LCT with n : (ln n)p ! 1). p1 (b) Converges (Use the LCT with n n ). p1 (c) Diverges (Use the LCT with n ). 1 1 1 (d) Converges (Use the comparison test: n2−ln n ≤ n2−n ≤ n(n−1) ). (e) Converges (Use the comparison test: 1 ≤ 1 as ex ≥ x). en2 n2 (f) Diverges (Use the LCT with 1 : n ! 1). n 1+ 1 n n tan 1 sec2( 1 )(− 1 ) (g) Diverges (Use the LCT with 1 : lim n = lim n n2 = 1): n n!1 1 n!1 − 1 n n2 π 2 1 1−cos n π (h) Converges (Use the LCT with n2 : 1 ! 2 ). n2 (ln n) sin 1 sin 1 (i) Converges (Use the LCT with p1 : n2 = lnpn n2 ). n n p1 n 1 n n n2 −1 π tanp p2 (j) Converges (Use the comparison test: n n ≤ n n ). 1 1 P1 1 (k) Diverges because (n + 2)(1 − cos n ) ≥ n(1 − cos n ) and n=1 n(1 − cos n ) diverges: 1 1 n(1−cos n ) 1−cos n 1 1 = 1 ! 2 . n n2 3+cos n 4 1 n (`) Converges (Use the comparison test: 0 ≤ en ≤ en = 4( e ) ): 3 (m) Converges because both P1 2 and P1 sin (n+1) converge. n=1 2n+n2 n=1 2n+n2 p p n+1− n 1 p 1 p 1 (n) Converges because n = n < 3 . n+1+ n n 2 14. (a) Note that, for any p 2 , jS − S j ≥ an+1+an+2+ ::: +an+p = An+p−An ! 1 as N n+p n An+p An+p p ! 1. (b) Take b = 1 . n An.