Practice Problems 12 : Comparison, comparison and Cauchy condensation tests

P∞ 1. Let an ≥ 0 for all n ∈ N. If n=1 an converges then show that

P∞ 2 (a) n=1 an converges. Is the converse true ? P∞ √ (b) n=1 anan+1 converges. √ P∞ an (c) n=1 n converges. n P∞ an+4 (d) n converges using comparison or . n=1 an+5

P∞ 1 2. Let (an) be a such that an > 0 for all n and an → ∞. Show that n n=1 an converges.

P∞ P∞ P∞ 2 3. Let n=1 an be a convergent . Show that n=1 |an| diverges if n=1 an diverges.

P∞ a1+a2+ ... +an 4. Let an > 0 for all n ∈ N. Show that the series n=1 n diverges. P∞ P∞ 5. Assume that n=1 an and n=1 bn are convergent and an, bn ≥ 0 for all n ∈ N. Show P∞ p 2 2 that n=1 an + bn converges. Does the converse hold ? P∞ 2 P∞ 2 P∞ p 6. Let an, bn ∈ R for all n and n=1 an and n=1 bn converge. Show that n=1(an − bn) converges for all p ≥ 2.

P∞  nπ  7. Show that n=1 sin n+1 diverges.

3 2 P∞ √an 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 P∞ a converges. Show that the series P∞ 1 − sin an n n=1 n n=1 an converges.

P∞ 1 1 10. Consider the series n=1 an where an = n for n = 1, 4, 9, 16, ... and an = n2 otherwise P∞ (i.e., if n is not a perfect square). Show that n=1 an converges but nan 9 0. P∞ 11. Let (an) be a sequence of positive real numbers such that an+1 ≤ an for all n and n=1 an P∞ converge. Show that n=1 n(an − an+1) converges. P∞ 1 12. Show that n=4 n(ln n)(ln(ln n)) diverges. P∞ 13. In each of the following cases, discuss the convergence/ of the series n=2 an where an equals: 1 sin( ) 2 1 √ n ln√n 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) tan√ n 1+ 1 n n 2 n n n n n √ √ 1 3+cos n 2+sin3(n+1) n+1− n (k) (n + 2)(1 − cos n )(`) en (m) 2n+n2 (n) n P∞ 14. (*) Suppose that an > 0 for all n and n=1 an diverges. Let (An) be the sequence of partial sums of P∞ a and (S ) be the sequence of partial sums of P∞ 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 P∞ 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 . √ 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→∞ an+5 4 1 1 2. Observe that n < n eventually. an 2 2 3. Since an → 0, an ≤ |an| 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. P∞ 2 p 2 6. It is sufficient to show that n=1(an − bn) converges because |an − bn| ≤ (an − bn) P∞ 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 √ 1 √an 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 P∞ 1 bounded above by ( 22 + 32 + 52 + ...) + (1 + 4 + 9 + ...) ≤ 2 n=1 n2 but nan = 1 when n is a perfect square. P∞ 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 (b) Converges (Use the LCT with n n ). √1 (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 n→∞ − 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 √1 : n2 = ln√n n2 ). n n √1 n 1 n n n2 −1 π tan√ √2 (j) Converges (Use the comparison test: n n ≤ n n ). 1 1 P∞ 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 P∞ 2 and P∞ sin (n+1) converge. n=1 2n+n2 n=1 2n+n2 √ √ n+1− n 1 √ 1 √ 1 (n) Converges because n = n < 3 . n+1+ n n 2 14. (a) Note that, for any p ∈ , |S − S | ≥ an+1+an+2+ ... +an+p = An+p−An → 1 as N n+p n An+p An+p p → ∞. (b) Take b = 1 . n An