① (al
I,
in,
if HEI 't II.I
-
- i E. LET
This is a geometricseries .
Recall that §oZ " conversesiff IZ la }
with sum equal to¥2 if it
converges .
Here wehave.
E- I and let= lil = Isl .
Thus, I ,converges
to the sum
Ei÷.
-
-
i Edit: i= i
n-
- i =T
① (b) We have that
⇐ 9n+, Yet ,
'+3
= let it [¥6t 2¥, t ÷g
t . c)
= Ceti ) a [Itt at . . .]
-
- teal¥. ⇐ 2¥= cent¥. I!⇒
"
Notethat I#120.16<1 .
Thus,this geometric
series convergesto
let it¥. . ( i÷⇒=kHz÷GI
① (c)
Note that
⇐ Yr¥ st ,t÷st "
= fit '÷r,.tt#rsztiJ
= E.
i= ⇐ KEY
Note that 12%-1=2.8971.
Thus,
the above geometricseries
diverges .
① (d) Note that ,
his:L 's'ii÷¥÷÷⇒Also ,
df¥him.
hittin thin.fi#m--hi;oitiTn--hin.En=oThus, lignin -- o .
.iSo, him.
D= # = ' to .
So, by
the divergencetheorem , £
,
n
diverges .
① let
we need to use partial fractions here .
Let's solve
n¥= htt ¥,
which becomes
I = A Intl )t Bn
(*)
This mustbe true
for all n .
Plug in n=- I into
CH to get
I = A ColtBC- l l
B = - I
Plug inn=o
into 1*1 to get
I = All )t B (o )
Atl .
Thus,
n¥= ht - htt t - n > l.
Let 's lookat the partial
sums
s.
-
- E.in#t=E..Lnt-⇒We have that
s, -4 - E)
=L - E
'Ii
sit'÷¥ 'tHt 's
s.-
- H -Htt't# H- tht 't
:
In general
Sh = I - htt
So, k l
his. Eo= Lissa
=
-
- hi:( that
= I - O
= I
Thus, &,¥,converses
and
⇐ ninth = I.
② If no =L,the series are the same .
Suppose he > l .
Let she a ,t . . - t an
be the partial
sumsof ⇐ an and
I
Sh'= an . tarot,t - - -t an.tk
be the partial
x
sums ofSan .
n-
- he
⇐D) Suppose ⇐ an exists.
Then,
Iim Snes forsome SE
Q .
Ktx
Note that
Sn .+u=A,tact - eat an
.- Itanotan.tt .
-it Anon
= Ait Azt. . . tano- ,t SI
= wt SI
where w-
- a ,tart . . - tan .- ,
is a fixed
complex number .
So,
hip.
si -- his.tn#-wl--fhi;nsno+a ) - w= S-W .
Thus, Eman exists
and
w
-
€n.
an = €,
an- ( a. tazt.cat an . . . ) .
x
⇐ Now supposeEan exists .h= no
Then , Lin,Su'-_ s' fo- some
SEE.
As before wehave Sn .+k=
wtsu .
Thus, high. sn.tn
-
- Lingo (Wtsi )
= W thingy Sn'= wts !
Thus, tiny, Su = tiff snotu
= w t s'.
So, ⇐ an converges
and
re
⇐ an = wt s'= a
,t . . . tano , t
E an
n = neo
③Let n
s =[ an = a , t azta,t ay t . . .
n k -- I
andn
Sh'= E bae = b
,t bet b, t by
t . . .
b- I
be the partialsumsfr the two series.
Then lim Sn -- A and 1in Sn'= B
.
htxthx
x
(at
The partial sumsfor the
series [(ahtba)b.=L
are n
si'-
- ⇐ Cant but =ant be sntsi .
I
Thus,
1in Sn"-
- high thingAtB
[propertyofconuergentseguen#
So,{ (aatbu ) converges to At B .
K-- Ix
(b) The partial sums of { Laware
G- i
Sn"- E. Kaul -- a an
-
- as.
Thus,
hip.si"- Ling knifed him
.
"
pr;gq7en=L A .
So, ⇐Law converges
to LA .
④ n
Let Sn -_ aitazt . . . tan= ? an denote
k l
the n - th partial sum of the series.
⇐8) Suppose that €, akconverges .
Then ( sin ,
converges .
Thus , Csn): ,is a Cauchy sequence
.
Let E >0 .
Then since ( Snl isa Cauchy sequence
there existsN > 0 where
it
n,m > N then Isn. - Sn ICE ,
Ctl
Let nz Nand m=ntp
where pzl .
ntp
Then,m > N
also and
Sm - Sn-_ Smp
- sn=,!ak - § ,ak= { 9kk=nH
ntp
So Lt ) gives / Earl CEK'- htt
(G) Suppose that for every E > 0
IN > o so thatif n> N then I ! .ae/ce
for p-
-I,2,3, . . .
We can use thisto show that
(Snh? ,
is aCauchy sequence .
Let e >o .
Then fromour
assumptionthere
exists N > 0
where itn> N
then I II!.ae/aE
for 13=1,43, ' - -
Let n,m > N .
Without lossof generality suppose
m >n .
Then m=ntpfor some p > I .
'" ism-snt-IE.iaa-E.in/--lEiiinteSo,( Snl? ,
is Cauchy . x
Thus, (g)I , converges
.
Hence §,akconverges. ⑤
⑤ Let Sn be the partial sums
x
of Eau and Sn' he the partial sums
herx
of Ebu '
k#(a) Suppose
Ebr converges .
Let E >0 .
By theCauchy criterion
for series (problem 4)
there existsN 70 so
that if
n > Nthen
bntitbntzt . - itbn+p¥/bn+ ,tbntzt . . -
t bnp / CE
bkreajerno.sifor all p >
I .
Since OcalaElse for all k
we getthat
/ ant , t antzt . . .t antp I = anti tant zt
. . . tant p
E bntitbntztiactbntpCE
for all p > I . Thus, bythe Cage hy
criterion for series ( problem 4) , Eakconverges
K -- i
x
(b) Suppose that Saia diverges .
K-- I
Thus, Csn)? ,
diverges .
Note that since each ak> 0, the
sequence Sn = a. tact. - it an is
an increasing sequence .
If (Snl-
wasbounded , then
by the
n= ,
monotone convergencetheorem in
real analysis ( 4650) ,the sequence
Sn would have alimit .
Thus , Csn )? ,
isunbounded , ie sniffy
.
Since akEbi. for all
k,this
tells usthat
Sn = a , tazt. ..t an
E b ,tbzt . -it bn
= s!
Thus , thesequence
Sn' is also
unbounded,
ie si →xas ntx .
So,Csn' )n? ,
does not convergeand §.ba diverges
⑥ (a) Consider the sequence 7£,
sinful
Nole that
F-=
'''
- et)sin (Iti)
=
E- "til e-icai ,
= (E-E) to
sink -it -- Sint-TKO = Effie'T -tosin Cri't =
sin C- ti) F- sin (Ii)
4¥%¥EsSin (ti
" ) = sinful= O
sin Crist -_sinHil=¥Gisin ( it e-
61£ sin Cti') = O
! ! The termsalternate between
the abovefour numbers
and
hence don't goto 0 . By
the divergencethan,this
series diverges.
⑥ Cbl
Note that 0jinewalI "n f-
'Y"tY;=INZ
K-ifnt-I-ilh.TLAnd , £
,
# = 2€,
ht converges .
Thus, ⇐ /ltn
"
/ converges bythe
comparisontest since I
' IE F- for
all n .
(problem 5)
So, €,
'converges
absolutely.
⑥ (c) Suppose 12-1<1 .
Then, Ei.IE/--nE...IzIn=IzltIzlZHZ/'t ' ' ' I
=IzI[ltlzltlzl't §# Iz ,
fwmc
Thus, €
,
Zn convergesabsolutely
if IZKI .
-⑥ (d) Suppose I -2171 .
Then high't- hi.gg/zIn--x
.By the
divergencetest , I, III
diverges .
Thus, £7
,
Z" does not converge
uniformly if17131 ,
⑦
Let Sn = §,
ah be the n-th partial
sumof the series .
Since ⇐ an converges, Ling,Sn = S
for someSE Cl .
Thus,
Iim an= hi,mx((qtazt - - -
tan) - (aitazt. .# any)
ht
= him. Csn- su]
= line,Sn- Inigo
Sn - i
re
=s - s
=O
⑧ ( al
Consider the K - th partial sum
Sh -- T t It f-too . t tf
we will showthat (Sula? , is
unbounded and hence cannotconverge .
We lookat a sub - series .
Note that
{.-- S
,=/
Sz , = Sz=It I
sa-
- S,-_ It Etf's tf )
> it 'zt¥ )Z
= It It I = It 2. I
Sz, = It It ( ft 4) + ( ft ft Itf)
> It It ( Etty )+ ( ft ft Itf )
a- T
= It It I+ I = It 3. I
In general,
{n 7It k
. 'z
Thus, {wereask→x .
Therefore , ( she)E= ,is unbounded
and thus diverges .
So, ⇐ In diverges .
⑧ (b) Let PER with PII .
x
Casely Suppose pal .
Then €,
ht = §,
I
which diverges by 8cal.
Casely Suppose pal .
Then, ht > NI
for all n >l.
Thus,by the
comparisontest
since ⇐ I diverges,so does
http .