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MA 101 (Mathematics I)
Hints/Solutions for Practice Problem Set - 2Ex.1(a)State TRUE or FALSE giving proper justication: If (xn) is a sequence inRwhich
converges to 0, then the sequence (xnn) must converge to 0. Solution: The given statement is TRUE. Ifxn!0, then there existsn02Nsuch thatjxnj<12 for allnn0and so 0 jxnnj<(12 )nfor allnn0. Since (12 )n!0, by sandwich theorem, it follows thatjxnnj !0 and consequentlyxnn!0. Ex.1(b)State TRUE or FALSE giving proper justication: There exists a non-convergent se- quence (xn) inRsuch that the sequence (xn+1n xn) is convergent. Solution: The given statement is FALSE. If possible, let there exist a non-convergent sequence (xn) such that the sequence (yn) is convergent, whereyn=xn+1n xn= (1 +1n )xnfor alln2N.
Then, sincexn=yn1+
1n for alln2Nand since (1+1n ) converges to 16= 0, it follows that (xn) must be convergent, which is a contradiction. Ex.1(c)State TRUE or FALSE giving proper justication: There exists a non-convergent se- quence (xn) inRsuch that the sequence (x2n+1n xn) is convergent. Solution: The given statement is TRUE, because ifxn= (1)nfor alln2N, then (xn) is not convergent, but (x2n+1n xn) = (1 +(1)nn ) is convergent (with limit 1), since(1)nn !0. Ex.1(d)State TRUE or FALSE giving proper justication: If (xn) is a sequence of positive real numbers such that the sequence ((1)nxn) converges to`2R, then`must be equal to 0. Solution: The given statement is TRUE. Since (1)nxn!`, the subsequences ((1)2nx2n) = (x2n) and ((1)2n1x2n1) = (x2n1) of ((1)nxn) must also converge to`. Sincex2n>0 for alln2N, `0 and sincex2n1<0 for alln2N,`0. Hence`= 0. Ex.1(e)State TRUE or FALSE giving proper justication: If an increasing sequence (xn) in Rhas a convergent subsequence, then (xn) must be convergent. Solution: The given statement is TRUE. Let (xnk) be a convergent subsequence of (xn). Then (xnk) is bounded above,i.e.there existsM >0 such thatxnkMfor allk2N. For eachk2N, knkand since (xn) is increasing, we getxkxnkM. Thus (xn) is bounded above and consequently (xn) is convergent. Ex.1(f)State TRUE or FALSE giving proper justication: If (xn) is a sequence of positive real numbers such that lim n!1(n32 xn) =32 , then the series1P n=1x nmust be convergent. Solution: The given statement is TRUE. Since the sequence (n32 xn) is convergent, it is bounded and so there existsM >0 such that 0n32 xnMfor alln2N. Hence 0xnMn
3=2for all
n2N. Since1P n=1Mn
3=2is convergent, by comparison test,1P
n=1x nis convergent. Ex.1(g)State TRUE or FALSE giving proper justication: If (xn) is a sequence of positive real numbers such that the series 1P n=1n2x2nconverges, then the series1P n=1x nmust converge.
Solution: For eachn2N, we havenP
k=1x k=nP k=11k kxk(nP k=11k 2)12 (nP k=1k2x2k)12 (using Cauchy-
Schwarz inequality). Since both the series
1P n=11n
2and1P
n=1n2x2nare convergent, their sequences of partial sums are bounded. Hence the sequence nP k=1x k 1 n=1of partial sums of the series1P n=1x nis bounded above. Therefore by monotonic criterion for series, the series 1P n=1x nis convergent. Ex.1(h)State TRUE or FALSE giving proper justication: If (xn) is a sequence inRsuch that the series 1P n=1x3nis convergent, then the series1P n=1x4nmust be convergent. Solution: The given statement is FALSE. Ifxn=(1)nn
1=4for alln2N, then1P
n=1x3n=1P n=1(1)nn 3=4is convergent by Leibniz's test (we note that the sequence ( 1n
3=4) is decreasing and converges to 0),
but 1P n=1x4n=1P n=11n is not convergent. Ex.1(i)State TRUE or FALSE giving proper justication: If (xn) is a sequence of positive real numbers such that the series 1P n=1x3nis convergent, then the series1P n=1x4nmust be convergent.
Solution: The given statement is TRUE. If1P
n=1x3nis convergent, thenx3n!0. So there exists n
02Nsuch thatx3n<1 for allnn0. Hencexn<1 for allnn0and therefore 0< x4n< x3n
for allnn0. Since1P n=1x3nis convergent, by comparison test,1P n=1x4nmust be convergent. Ex.1(j)State TRUE or FALSE giving proper justication: If (xn) is a sequence of positive real numbers such that the series 1P n=1x4nis convergent, then the series1P n=1x3nmust be convergent.
Solution: The given statement is FALSE. Ifxn=1n
1=3for alln2N, then1P
n=1x4n=1P n=11n 4=3is convergent, but 1P n=1x3n=1P n=11n is not convergent. Ex.1(k)State TRUE or FALSE giving proper justication: Iff:R!Ris continuous at both 2 and 4, thenfmust be continuous at somec2(2;4). Solution: The given statement is FALSE. Letf(x) =(x2)(x4) ifx2Q;
0 ifx2RnQ:
Let (xn) be any sequence inRsuch thatxn!2. Sincejf(xn)j j(xn2)(xn4)j !0, f(xn)!0 =f(2). This shows thatf:R!Ris continuous at 2. Similarlyfis continuous at 4. Letc2(2;4). Then there exist sequences (rn) inQand (tn) inRnQsuch thatrn!candtn!c. Sincef(rn) = (rn2)(rn4)!(c2)(c4)6= 0 and sincef(tn)!0, it follows thatfcannot be continuous atc. Ex.1(l)State TRUE or FALSE giving proper justication: There exists a continuous function f:R!Rsuch thatf(x)2Qfor allx2RnQandf(x)2RnQfor allx2Q. Solution: The given statement is FALSE. If possible, let there exist a continuous functionf:R! Rsuch thatf(x)2Qfor allx2RnQandf(x)2RnQfor allx2Q. Letg(x) =xf(x) for allx2R. Theng:R!Ris continuous andg(x)2RnQfor allx2R. By the intermediate value theorem, it follows thatgmust be a constant function. Henceg(x) =g(0) for allx2Rand sof(x) =x+f(0) for allx2R. In particular, we getf(f(0)) = 2f(0), which is a contradiction, sincef(0) =g(0)2RnQ. Ex.1(m)State TRUE or FALSE giving proper justication: Iff: [1;2]!Ris a dieren- tiable function, then the derivativef0must be bounded on [1;2]. Solution: The given statement is FALSE. Letf(x) =(x1)2sin1(x1)2if 1< x2;
0 ifx= 1:
Clearlyf: [1;2]!Ris dierentiable on (1;2] withf0(x) = 2(x1)sin1(x1)22x1cos1(x1)2for allx2(1;2]. Also, sincef(x)f(1)x1=jx1jjsin1(x1)2j jx1jfor allx2(1;2], it follows that lim x!1+f(x)f(1)x1= 0 and hencefis dierentiable at 1 (withf0(1) = 0). Ifxn= 1 +1p2nfor all n2N, thenxn2[1;2] for alln2Nandf0(xn) =2p2n! 1, which shows thatf0is not bounded on [1;2]. Ex.1(n)State TRUE or FALSE giving proper justication: Iff: [0;1)!Ris dierentiable such thatf(0) = 0 = limx!1f(x), then there must existc2(0;1) such thatf0(c) = 0. Solution: The given statement is TRUE. If possible, letf0(x)6= 0 for allx2(0;1). Then by the intermediate value property of derivatives, eitherf0(x)>0 for allx2(0;1) orf0(x)<0 for allx2(0;1). We assume thatf0(x)>0 for allx2(0;1). (The other case is almost similar.) Thenfis strictly increasing on [0;1) and sof(x)> f(1)> f(0) = 0 for allx2(1;1). This contradicts the given fact that limx!1f(x) = 0. Hence there existsc2(0;1) such thatf0(c) = 0. Ex.1(o)State TRUE or FALSE giving proper justication: Iff:R!Ris dierentiable, then for eachc2R, there must exista;b2Rwitha < c < bsuch thatf(b)f(a) = (ba)f0(c). Solution: The given statement is FALSE. Letf(x) =x3for allx2R, so thatf:R!Ris dier- entiable. If possible, let there exista;b2Rwitha <0< bsuch thatf(b)f(a) = (ba)f0(0). Thenb3a3= (ba)0 = 0)b3=a3, which is not true, sincea <0 andb >0. Ex.1(p)State TRUE or FALSE giving proper justication: The functionf:R!R, dened by f(x) =x+ sinxfor allx2R, is strictly increasing onR. Solution: The given statement is TRUE. Sincef0(x) = 1+cosx0 for allx2R,fis increasing onR. If possible, let there existx1;x22Rwithx1< x2such thatf(x1) =f(x2). Thenfmust be constant on [x1;x2] and sof0(x) = 0 for allx2[x1;x2]. This implies that cosx=1 for all x2[x1;x2], which is not true. Thereforefis strictly increasing onR. Ex.1(r)State TRUE or FALSE giving proper justication: Iff: [0;1]!Ris a bounded function such that lim n!11n n P k=1f(kn ) exists (inR), thenfmust be Riemann integrable on [0;1]. Solution: The given statement is FALSE. Iff(x) =0 ifx2[0;1]\Q;
1 ifx2[0;1]\(RnQ);
thenf: [0;1]!Ris a bounded function and we know thatfis not Riemann integrable on [0;1].
However, sincef(kn
) = 0 fork= 1;:::;nand for alln2N, limn!11n n P k=1f(kn ) = 0.
Ex.2(a)For alln2N, letan=n+1n
andxn=1n
2(a1++an). Examine whether the
sequence (xn) is convergent. Also, nd the limit if it is convergent.
Solution: For alln2N,xn=1n
2[(1+2++n)+(1+12
++1n )] =12 (1+1n )+1n 1+12 ++1nn . Since 1n !0, by the solution of Ex.4 of Practice Problem Set - 2, we get1n (1+12 ++1n )!0. It follows (by limit rules for algebraic operations) that (xn) is convergent with limit12 (1 + 0) + 0:0 =12
Alternative solution: We can show that limn!11n
2(1 +12
++1n ) = 0 even without using Ex.4 of
Practice Problem Set - 2. We have 01n
2(1 +12
++1n )1n
2(1 ++ 1) =1n
for alln2N. Since 1n !0, by sandwich theorem, it follows that1n
2(1 +12
++1n )!0.
Ex.2(b)Letxn= (n2+ 1)18
(n+ 1)14 for alln2N. Examine whether the sequence (xn) is convergent. Also, nd the limit if it is convergent.
Hint: We havexn= (n2+ 1)18
(n2)18 +n14 (n+ 1)14 for alln2N. Now consider the rst two terms together and the last two terms together. The limit is 0.
Ex.2(c)Letxn= (n2+n)1n
for alln2N. Examine whether the sequence (xn) is conver- gent. Also, nd the limit if it is convergent.
Solution: We have 1xn(2n2)1n
for alln2N. Since 21n !1 andn1n !1, it follows that (2n2)1n = 21n (n1n )2!1. Hence by sandwich theorem, (xn) is convergent with limit 1.
Ex.2(d)Letxn= 5n(1n
31n!) for alln2N. Examine whether the sequence (xn) is conver-
gent. Also, nd the limit if it is convergent.
Solution: Letan=5nn
3andbn=5nn!for alln2N. Since limn!1jan+1a
nj= limn!15(1+ 1n )3= 5>1 and lim n!1jbn+1b nj= limn!15n+1= 0<1, the sequence (an) is not convergent and the sequence (bn) is con- vergent (with limit 0). Since (xn) = (an)(bn), it follows that (See Ex.1(c) of Practice Problem
Set - 1) (xn) is not convergent.
Ex.2(e)Letxn=11:n+12:(n1)+13:(n2)++1n:1for alln2N. Examine whether the se- quence (xn) is convergent. Also, nd the limit if it is convergent.
Solution: We havexn=1n+1[(1 +1n
) + (12 +1n1) ++ (1n + 1)] =2nn+11n (1 +12 ++1n ) for all n2N. Since1n !0,1n (1+12 ++1n )!0 (using the solution of Ex.4 of Practice Problem Set -
2) and
2nn+1=21+
1n !2. Hence by limit rule for product, (xn) is convergent and limn!1xn= 0.
Ex.2(f)Letxn=n3
[n3 ] for alln2N. Examine whether the sequence (xn) is convergent.
Also, nd the limit if it is convergent.
Solution: We havex3n= 0 andx3n+1=13
for alln2N. Thus (xn) has two subsequences (x3n) and (x3n+1) converging to two dierent limits,viz.0 and13 respectively. Therefore (xn) is not convergent. Ex.2(g)Letx1= 1 andxn+1= (nn+1)x2nfor alln2N. Examine whether the sequence (xn) is convergent. Also, nd the limit if it is convergent. Solution: Clearlyxn0 for alln2N. Also, we havex1= 1 and if we assume thatxk1 for somek2N, thenxk+1= (kk+1)x2k1. Hence by the principle of mathematical induc- tion,xn1 for alln2N. This givesxn+1= (nn+1xn)xnxnfor alln2N. Thus (xn) is decreasing and bounded below and hence (xn) is convergent. If`= limn!1xn, then we have lim n!1xn+1= limn!1nn+1( limn!1xn)2)`=`2)`= 0 or 1. Since`= inffxn:n2Ng x2=12 , we must have`= 0.
Ex.2(h)Leta;b2R,x1=a,x2=bandxn+2=12
(xn+xn+1) for alln2N. Examine whether the sequence (xn) is convergent. Also, nd the limit if it is convergent.
Solution: We havexn+1xn= (12
)(xnxn1) == (12 )n1(x2x1) for alln2N. Hence x n=x1+(xnxn1)++(x2x1) =a+[(12 )n2++1](x2x1) =a+23 [1(12 )n1](ba) for alln2N. Since (12 )n!0, (xn) is convergent and limn!1xn=a+23 (10)(ba) =13 (a+2b). Alternative solution: The convergence of (xn) can also be shown as follows.
We havexn+2xn+1= (12
)(xn+1xn) for alln2N, so thatjxn+2xn+1j=12 jxn+1xnjfor all n2N. Hence it follows that (xn) is a Cauchy sequence inRand therefore (xn) converges.
Ex.2(i)Let 0< xn<1 andxn(1xn+1)>14
for alln2N. Examine whether the sequence (xn) is convergent. Also, nd the limit if it is convergent. SolutionUsing theA:M: > G:M:inequality, we havexn+(1xn+1)2 px n(1xn+1)>12 for all n2N. Hencexn> xn+1for alln2Nand so (xn) is decreasing. Sincexn>0 for alln2N, (xn) is bounded below. Therefore (xn) is convergent. If limn!1xn=`, then limn!1xn+1=`. Since x n(1xn+1)>14 for alln2N, we get`(1`)14 )(2`1)20)(2`1)2= 0)`=12 Ex.3Let (xn) be any non-constant sequence inRsuch thatxn+1=12 (xn+xn+2) for alln2N.
Show that (xn) cannot converge.
Solution: For eachn2N, 2xn+1=xn+xn+2)xn+2xn+1=xn+1xn. Ifd=x2x1, then x n=x1+(n1)dfor alln2N. Since (xn) is not a constant sequence,d6= 0. Given anyM >0, choosingn2Nsatisfyingn >1 +M+jx1jjdj, we nd thatjxnj> M. Thus (xn) is unbounded and consequently (xn) cannot converge.
Ex.4Let (xn) be a sequence inRand letyn=1n
(x1++xn) for alln2N. If (xn) is convergent, then show that (yn) is also convergent. If (yn) is convergent, is it necessary that (xn) is (i) convergent? (ii) bounded? Solution: Letxn!`2Rand let" >0. Then there existsN2Nsuch thatjxn`j<"2 for alln >
N. Now for alln > N, we havejyn`j=1n
j(x1`)++(xn`)j 1nquotesdbs_dbs14.pdfusesText_20
[PDF] 447 HOMEWORK SET 6 1 33 3) Let x1 ≥ 2 and x n+1 := 1 + √ xn