[PDF] [PDF] Practice Problems 2: Convergence of sequences and monotone

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[PDF] Practice Problems 2: Convergence of sequences and monotone

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(1) Let (xn) be a sequence such that xn → x and xn → y bounded and monotone, and find its limit (a) x1 = 1 and xn+1 = 4+3xn 3+2xn (b) xn+1 = 1 2 ( xn + a

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Practice Problems 2: Convergence of sequences and monotone sequences

1. Investigate the convergence of the sequence (xn) where

(a)xn=11+n2+22+n2+:::+nn+n2: (b)xn= (an+bn)1=nwhere 0< a < b. (c)xn= (p2213 )(p2215 ):::(p2212n+1): (d)xn=n(n+ 1)for some2(0;1): (e)xn=2nn!: (f)xn=12+34++(1)n1nn

2. Letxn= (1)nfor alln2N. Show that the sequence (xn) does not converge.

3. LetAbe a non-empty subset ofRand= infA. Show that there exists a sequence (an)

such thatan2Afor alln2Nandan!.

4. Letx02Q. Show that there exists a sequence (xn) of irrational numbers such thatxn!x0.

5. Let (xn) be a sequence inR. Prove or disprove the following statements.

(a) Ifxn!0 and (yn) is a bounded sequence thenxnyn!0. (b) Ifxn! 1and (yn) is a bounded sequence thenxnyn! 1.

6. Let (xn) be a sequence inR. Prove or disprove the following statements.

(a) If the sequence (xn+1n xn) converges then (xn) converges. (b) If the sequence (x2n+1n xn) converges then (xn) converges.

7. Show that the sequence (xn) is bounded and monotone, and nd its limit where

(a)x1= 2 andxn+1= 21x nforn2N. (b)x1=p2 andxn+1=p2xnforn2N. (c)x1= 1 andxn+1=4+3xn3+2xn;forn2N.

8. Let 0< b1< a1and denean+1=an+bn2

andbn+1=pa nbnfor alln2N. Show that both (an) and (bn) converge.

9. Leta >0 andx1>0:Denexn+1=12

x n+ax n for alln2N:Show that the sequence (xn) converges topa:

10. Let (xn) be a sequence in (0;1). Suppose 4xn(1xn+1)>1 for alln2N. Show that the

sequence is monotone and nd the limit.

11. LetAbe a non-empty subset ofRandx02R. Show that there exists a sequence (an) in A

such thatjx0anj !d(x0;A). Recall thatd(x0;A) =inffjx0aj:a2Ag.

12. Let (an) be a bounded sequence. For everyn2N;denexn= supfak:k > ng. Show that

the sequence (xn) converges.

13.(*)Show that the sequence (en) dened byen= (1+1n

)nis increasing and bounded above.

Practice Problems2: Hints/Solutions

1. (a) Since (1 + 2 +:::+n)1n+n2xn(1 + 2 +:::+n)11+n2,xn!12

(b) Note thatb= (bn)1=nxn(2bn)1=n= 21=nb!b:Thereforexn!b. (c) We have 0< xn<(p21)nand hencexn!0. (d) Observe thatxn=n[(1 +1n )1]< n[1 +1n

1] =1n

1!0. Hencexn!0.

(e) Consider xn+1x nand apply the ratio test for sequences to conclude thatxn!0. (f) Herex2n=12 andx2n+1=n+12n+1!12 . The sequence does not converge.

2. Supposexn!x0for somex0. Then, by the denition, for=14

(why14 ?) there existsNsuch thatjxnx0j<14 for allnN. Then for allm;nN;jxnxmj jxnx0j+jxmx0j 14 +14 =12 which is not true becausejxnxn+1j= 2 for anyn.

3. Since+1n

is not a l.b., ndan2Asuch thatan< +1n . Allown! 1.

4. Find an irrationalxnsatisfyingx0< xn< x0+1n

for everyn2N. Allown! 1.

5. (a) True. FindM2Nsuch that 0 jxnynj< Mjxnj. Allown! 1.

(b) False. Takexn=n2andyn=1n

6. (a) Letyn=xn+1n

xn= (1+1n )xn:Thenxn=yn(1+ 1n ):Hence (xn) converges if (yn) converges. (b) The statement is not true. Take, for example,xn= (1)n.

7. (a) Observe thatx2< x1. Ifxn< xn1, thenxn+1<21x

n1=xn. The sequence is decreasing. Note thatxn>0. The sequence converges and the limit is 1. (b) Observe thatx2> x1. Sincex2n+1x2n= 2(xnxn1), by induction (xn) is increasing. It can be observed again by induction thatxn2. The limit is 2. (c) Note thatx2> x1. Sincexn+1xn=xnxn1(3+2xn)(3+2xn1), by induction (xn) is increasing.

Note thatxn+1= 1 +1+xn3+2xn2. The limit isp2.

8. By the AM-GM inequalitybnan. Therefore 0an+1an+an2

=an. Note that b n+1pb nbn=bnandbnana1. Use monotone criterion for both (an) and (bn).

9. Note thatxn>0 andxn+1xn=12

(xn+ax n)xn=12 (ax2nx n):Further, by the A.M -G.M. inequality,xn+1pa:Therefore (xn) is decreasing and bounded below.

10. By the AM-GM inequality

xn+(1xn+1)2 px n(1xn+1)>12 . Thereforexn> xn+1. Supposexn!x0for somex0. Then 4x0(1x0)1 which implies that (2x01)20.

Thereforex0=12

11. Use Problem 2 or follow the steps of the solution of Problem 2.

12. Observe that the sequence (xn) is decreasing and bounded.

13. By binomial theoremen= 1+1+12!

(11n )+13! (11n )(12n )+:::+1n!(11n ):::(1n1n )en+1 anden2 +12! +13! +:::+1n!2 +12 +12

2+:::+12

n13: Alternate Solution:For eachn2N, apply AM-GM inequality fora1= 1;a2=a3=::::= a n+1= 1 +1n . We geten+1> en.quotesdbs_dbs7.pdfusesText_5