Math Real Analysis

MATH 3001 W21 A2

D ue o

n Fe b 15

: I Show that Idnt - th) diverges .

A- I

(2) Find the sum of 2nF , n'+13ha i

Does E en n

ki n't4 converge ?

(4) Suppose I , an converges with each an

>o and set sn=F! ,

ax .

Show that I ,

an Sh and II , Afn both converge .

(5) Show that ft , zntzn. ,, converges and has the same sum

as the alternating harmonic series .

(6) Show that the Cauchy product of fro absolutely convergent series

converges absolutely .

H Find the sum of II. Fn . Decide on convergence ( absolute convergence( divergence :

Cal ⇐ C- if " b"" @ , E 3 "

n !

h n> , n

"

• 1 I

(b) I s- (f) I n't 1h41 nzz Cnn)

em

a 1

(g I sin th N Fy, Cfnnylnlehn)heel •

n ! A I

(d) I - Ch) I hittin h= , hh n I

MATH 3001 W21 S2

fatigue (htt-th ) (fntttsn) y

l" fntttfn

=

fnttt ST > 24nF '

since I , gnttt diverges, the result follows from the comparison test .

(2) n't"3nt2 = Cheikh-14 = n'ti - n'+2 = 9h

m

so Am -

- Ian -- ⇐E) + Its -41-114 -f) t - - i -4¥ - ha) HII

ha h=2 A-3 h=m

men = ta -M¥2 > L .

(3) fnre it does ! We have bn " ""50

, so In

, sit . buns n

"'

n'12

Hn>he .

Thus

fhh n'12 will < s = n

- "2

, twine

.

n't 4 n44 h '

the companion test shows convergence .

④ We have nigh sn= A = Eiga . So #host . Fn>no, Az ssn<2A (Note : Aso.) . So An>no : anSns 2Aan and offs Ian . In both cases, the companion test shows convergence .

⑤ 2h fin-it = In-i - Ln . So this send is the same as ¥, C-'th"

with h

brackets : C- ta ) + ( 'g - t ) t ( f- f)t . . . = I - tatty -ty tf -ft. . . h-4 h=2 n=3 N

add brackets

Since E ,

t't ""

converges, then the bracketing does not alter the convergence (nor the value) of the series .

(6) In an , II.G both absolutely convergent. on = It. ajfnj .

HE has

Cim = E. lent ' II E.o lait Ibn-it = laellb.lt (Keller, It 1911 thot)

+ dad lent-11941914 ladKrol) n=@ t- -it (laollbmlt 19,11hm.. It . .. -1 I am, I 1h It 19mL Ibd)

⇐ (Holt .-it 1am1) ( Ibolt . . - t Hml) = ftp.t.la; t ) ( E! If;D ⇐ AB

.

So Cim is increasing and founded above, hence it converges .

H II Tn = Fonts ) " = 3 E.on H )

"" = 3 ME

,

(m-if} ) "

=3 NE, n HY - 3 E

,

"

.

-W -

Eon " = 'k

So we get n "

= 314 .

(8) ⑨ converges by the alternating genes test . However, FLI > In th>2, so the absolute series diverges . This is thus a conditionally convergent series .

nm

ex yn÷ 7 gin = I '" this p genes with p

-

- 43 St . Diverges .

an = sink )

"

To . How quickly ? Recall : ¥n, sing =L, so

nhjn.si?Ynhl-- 1 . By the companion test, the send diverges .

④ an -- hunt ; "

any = I In. "

n ! = 4th anti: hey -- ⇐ it

"

= ( i - ha ) " "→°

> e- ' a # s 1

So the series converges ( ratio test)

an= "

nnn !

: math = 3 . ft- ha, ) " "T Ze >I as es3 .

So the send diverges .

H Integral test : y -

-thx y 's = eyeing

j dx t

! eyty dy tf ,

e-Hem -"

dy . New 2 Hh x)

thx =

buy > 2 for y > eh, so the integrand is founded above by e- Y which

is integrable at y→a. Hence the semis converges .

" I caxsdiien ; '

II es! %:: . eaten'" ay

ylny = elleny ) @ny)

New y - Hn y)

' > tag for y large enough . (Bk y > 2 Hny))

→ So the integral diverges ( et

' is not integrable since yhjm.fqyp.to at y →a) . Hence the Senor diverges .

Ch) an= n'it in . The intuition is that It 'In ~ 1 For large n and so the series should diverge (as NE, th does) . Set bn

= th .

FF Fn - ten. Fyn = needy n

- "n - -

Effy e- them

= e .

So by the limit comparison test, the send diverges .