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 .