Sequencias e Series

5/20/2018 Sequencias e Series

1/35


2/35


3/35

a

a+

a+ > a

a+

a

a

a

0+

0

1(0)2

= 10+

=

+ = =

=

1 = 0

+

10+

= +c=

a >0

a

=

a =

a >1

a =

10 00 0 0 1

0< a

a = 0+

0< b


4/35

xn n

x(n) n

1n

nN=

1, 1

2, , 1

n,

xn=

1n

n >0

limn

xn=L > 0 N0 N n > N0 = |xn L| <

L

|xn L| < L < xn < L+

limn

xn= M R N0 N n > N0 = xn > M

limn

xn=

M R

N0 N n > N0 = xn < M

xn xn

(an) (bn)

limn

(an+bn) = limn

an+ limn

bn

limn

(an) = limn

an

limn

(anbn) = limn

an limn

bn

limn

anbn

=limn

an

limn

bn

limn

bn= 0


5/35

f limn

f(xn) =f

limn

xn

ab = elna

b

= eb lna

a > 0

n

yk =xnk

knk

ni = nj i= j

xn yk xn

xn = (1)n x2n = (1)2n = 1 1

x2n+1= (1)2n+1 = 1 1

xn

(xn) (xn)

xn =

(

1)n

n

0

yk = x2k =(1)2k

2k = 1

2k

limk

yk = limk

1

2k =

1

= 0+ = 0

zk =x2k+1 =

(1)2k+12k+1

= 12k+1

limk

zk = limk

12k+ 1

=1 = 0

= 0

yn zn

yk zk (xn) xn 0

xn = (1)n yk = x4k = (1)4k+2 = 0

limk

yk = limk

1 = 1 zk = x4k+2 = (

1)4k+2 = 1

lim

kzk = lim

k1 = 1

yk zk

xn 1

xn


6/35

xn = f(n) f x lim

xf(x)

limn

xn = limx

f(x)

limn

en

n2 =

LHopital= lim

nen

2n=

LHopital= lim

nen

2 =

2

=

n

xn= sen(n)

xn= 0 lim

nxn= 0 lim

xsen(x) = sen() =

an bn cn lim

nan = lim

ncn= L lim

nbn=L

limn

an=L > 0 Na N n > Na L an L+ Nc N

n > Nc L cn L+ N = max{Na, Nc} n > N

L an L+ L cn L+ L < an bn cn< L+ limn

bn=L

xn =

cosnn

1 cos n 1

1n cosn

n 1

n

limn

1n

limn

cos n

n lim

n1

n

1 = 0 limn

cos n

n 1 = 0

limn

cos n

n = 0

limn

xn= 0 limn

|xn| = 0

xn=

(1)nn

lim

n

(1)nn = limn 1n = 1 = 0+ limn (1)

n

n = 0

limn

n!nn

= 0

n!nn

= n(n1)(n2)21nn

(n1) n n n

nn =

nn1

nn = 1

n

0 n!nn

1n

0 lim

nn!

nn lim

n1

n=

1

= 0 limnn!

nn = 0


7/35

yn xn

xn=o(yn) lim

nxnyn

= 0

f=o(g) a limxa f(x)g(x) = 0

limx

f(x) =

c= o(f)

a < b

xa =o(xb)

u >0 a > 1 logax= o(xu) xu =o(ax)

lnx

ex

ax = o((x)) (x) = o(xx) (n) = (n 1)! n

an =o(n!) n! =o(nn)

an = o(n!)

n! = o(nn)

x

f g f=o(g) lim

nf(n)

g(n) = 0

c logan xu

an

n! nn

u > 0

a > 1

a

e

1

na nb

an bn

a < b

f=o(g)

a

lim

xa(f(x) +g(x)) = lim

xag(x)

f = o(g)

a

lim

xaf(x)

g(x) = 0

limxa

(f(x) +g(x)) = limx

f(x)

g(x)+ 1

g(x) = lim

n(0 + 1) lim

xg(x) = lim

xag(x)

limn

1+n2en + ln n

1 =o(n2)

ln n= o(en)


8/35

limn

1+n2

en + ln n = lim

n

n2

en = lim

nn

en =

LHopital= lim

n1

en =

1

e =

1

= 0.

an an+1 an n

an an+1 an n

an+1 > an n

an+1 < an n

M

n, |xn| M

xn+1 =

nn+3

|xn| 1 xn+1xn n+1n+2 nn+1 (n+ 1)2 n(n+ 2) n2 + 2n+ 1 n2 + 2n 1 0

limn

xn = 1

xn+1 =f(xn)

L= lim

nxn L= f(L)

xn+1 =

x2n+22xn

L= lim

nxn

limn

xn+1 = limn

x2n+ 2

2xn=

limn

xn

2+ 2

2 limn

xn

L= L2+22L

2L2 =L2+2

L2 = 2 L= 2

limn

xn=

2

limn

xn =

2

x0 > 0

xn>0 L 0 x0>0 L=

2


9/35

limn

nn= 1

limn

nn= limn

eln(nn)

ln ( nn) = ln n 1n = 1

nln n= lnn

n

ln

limn

n

n

= limn

ln n

n =

ln =

LHopital= lim

n1/n

1 = lim

n1

n=

1

= 0

ln

limn

n

n

= 0 limn

n

n = limn

eln(nn) =elimn ln(

nn) =e0 = 1

limn

n

a= 1

a >0

limn an =

0, |a| 1

1, a= 1

, a 1

a > 0

ln a

a = 1

a = 0

a =1

a < 1

a >1

limn

n

n! = n! =n(n 1)(n 2) 2 1 n2 n

2 n2

n2 1! n

2

n2

n

n!

nn2

n2 =

n

2

limn

n

n!

limn

n

2=

2

=

limn

nn! =


10/35

an

n0 an an

an

n=0

(1)n = 1 1 + 1 1 +

n=0

(1)n = (1 1) + (1 1) + = 0 + 0 + = 0

n=0

(1)n = 1 + (1 + 1) + (1 + 1) + = 1 + 0 + 0 + = 1

n=n0

an=S SN=N

n=n0

an = an0+ aN

Sn0 = an0 SN = SN1+ aN N > n0

n=n0

an=S

limn

Sn= S

Sn Sn

n=0

(1)n

n=0

(1)n =12


11/35

n=0

(1)n = 1 1 + 1 1 +

S2k= (1)0 + + (1)2k = 1 1 + 1 1 + + 1 = (1 1)+(1 1) + + (1 1) + 1 =0 + + 0 + 1 = 1 lim

kS2k= 1

S2k+1 = (

1)0 +

+ (

1)2k+1 = 1

1 + 1

1 +

1 = (1

1)+(1

1) +

+ (1

1) =

0 + + 0 = 0 limk S2k+1 = 0

Sn

Sn n

an

bn

(an+bn) = an+ bn

(an) =

an

an |an| |an|

n

= 0

an

an

an lim

n|an| = 0

Sn = Sn1 +an

limn

Sn = limn

Sn1+ limn

an

limn

Sn=S S=S+ limn

an limn

an= 0

limn

|an| = 0

an

nn+1

limn

|an| = limn

n

n+ 1=

LHopital= lim

n

1

1= 1 = 0

(1)nn2(n+1)2


12/35

n=1

1

n(n+ 1)

1n(n+1)

=

an

+ bn+1

= a(n+1)+bnn(n+1)

= (a+b)n+an(n+1)

(a + b)n + a= 1

n

a+b = 0

a = 1

a= 1

b= a= 1

1n(n+1)

= 1n 1

n+1

Sn=

11 1

2

+12 1

3

+ + 1

n1 1n

+1n 1

n+1

= 1

1 1

n+1

n 1

n=1

1

n(n+ 1) = lim

nSn= lim

n

1 1

n+ 1

= 1 1 = 1)

n=1

ln

n

n+ 1

limn

an= 0

n=0 a0rn

a0

= 0

r

= 0

n=0

a0rn

a0= 0 |r| < 1

n=0

a0rn =

a01 r

n=0

a0rn

a0

= 0 r

= 0

a

n = a

n1r

n > 0

r

Sn = a0 + + an rSn = a0r + +anr = a1 + + an+1

Sn Sn rSn = a0 an+1 ai = a0ri (1r)Sn =a0 an+1 = a0 a0rn+1 = a0(1 rn+1) Sn = a0(1r

n+1)1r r= 1

|r| < 1

limn

|rn| = 0

limn

rn = 0

n=0

a0rn = lim

nSn = lim

na0(1 rn+1)

1 r = a01 r |r| 1 limn |a0r

n| = 0

limn

a0rn = 0

n=0

1

22n+3 =

n=0

1

22n1

23 =

n=0

1

8

14

n=

18

1 14

=1834

=1

6

a0 = 18

r = 1

4

|r|


13/35

1.23636 = 1.2+0.036+0.00036+ = 1.2+

n=0

0.036

102n

|r| < 1

a0 = 0.036 r =

1100

11.23636 = 1.2 + 0.036

11/100 =1210

+ 0.0360.99

= 1210

+ 36990

= 1188+36990

= 1224990

n=2

1

e3n+1 n 0 k =

n 2

n = 0

k = 0

n = k + 2

n=2

1

e3n+1 =

k=0

1

e3k+7 =

k=0

1

e7

1

e3

k=

1

e71

1 1e3

a0 =

1e7

r = 1

e3

|r|


14/35

y = f(x)

x

y

n n+ 1

anan+1

r

1

n=1(1)nn ln n

lim

nan = lim

n1

n ln n=

1

ln = 1

= 0

an=f(n) = 1n lnn f x f(x) = (ln x+1)(x lnx)2 0 x 1

0.05

an = 1n lnn

1n

n > e

1n 0.05 = n 20

n=0

(1)n(n 10.5)2

an an = f(n) f

f(x) 0

x

A

f(x)dx

A

S SN N

f(x)dx

n=n0

an= an0+ +aN+N

an=SN+

N+1

an. f(x) 0 n N

Sn n N

n=N+1

an

n > N

an+1 f(x) an x [n, n+ 1] f

an+1=

n+1n

an+1dx n+1n

f(x)dx n+1n

andx=an


15/35

n=N+1

an=aN+1+aN+2+ N+1N

f(x)dx+

N+2N+1

f(x)dx+ =N

f(x)dx

N

f(x)dx N+1N

f(x)dx+

N+2N+1

f(x)dx+ aN+aN+1+ =

n=N

an,

SN

n=N+1

an

limx

f(x) = 0

limn

an= 0

n

en2 an = f(n) f(x) =

x

ex2

x >0

f(x) = 1ex2xex22x(ex2)

2 = (1x2)ex2

e2x2 0 1< x2 x >1 x

x

ex2dx=

ex

2

xdx=1

2

ex

2

(2x)dx=12

ex2

+C=12ex2

+C.

1

x

ex2 dx= 12ex2

1

= 12e

2

12e1

2 =1

+ 1

2e=

1

2e

1n lnn

1np

p

p

1np

p >1

n=1

1n

n=0

1(n+ 1)3

=n=0

1

(n+ 1)3/2

k =n + 1

n= k 1

n=0

1

(n+ 1)3/2 =

k=1

1

k3/2

p

p= 32

>1

an bn an > 0 bn > 0 an bn

an=

bn =


16/35

bn an n=n0

an

n=n0

bn

(Sa)N=

Nn=n0

an (Sb)N=N

n=n0

bn

(Sa)n (Sb)n an limN (Sa)N= = limN

(Sa)N limN (Sb)N

bn =

bn (Sa)N (Sb)N limN (Sb)N=Sb< (Sa)N

Sb

(Sa)N (Sb)N = limN (Sa)N limN (Sb)N

n=n0

an

n=n0

bn

n=1

1

n(n+ 1)

1n(n+1

1n2

1n2

p

p= 2> 1

n=1

enn

= en

n

en

n en

en

r= 1

e0

= L2

N N

n > N = L < anbn

< L+

bn(L )< an 0

bn < an(L) anL bn

L= 0

bn = 1 N N n > N = 1 =

L < anbn

< L+ = 1

an < bn

bn

an

an

bn

an

bn lim

nanbn

=

an

bn

bn an

limn

anbn

= 0

limn

anbn

=

limn

bnan

= 0


17/35

1(n+1)2

an =

1(n+1)2

bn = 1n2

limnanbn = 1 = 0 bn = 1n2 p

p >1

1(n+1)2

n=0

en2+n

en

2

limn

anbn

=

an

|an| an

|an|

0 an + |an| 2|an| 2 |an| (an+ |an|)

an =

(an+ |an|)

|an|

cosnn2

|an| 1n2

1n2

p

p >1

|an|

r= lim n

|an|

r


18/35

r >1

an

r= 1

r

r= lim n|an| < 1 =

1r2

N

N

n > N

r < n

|an| < r+. r = r+ r < 1 |an| < rn

k=N

rk

|r| < 1

|an|

an

r= lim n

|an| >1 = r12 r= r rk < an

r >1

limn

an= limn

an= 0

en

n2 r = lim

nnenn2 = limn enn2 =e >1

limn

n

n2 = 1

r= lim

an+1an

r 1 an

r= 1

r= lim

an+1an N

r 1

limk

aN+k= limn

an= 0

r

r

r = 1

r


19/35

r

r = 0

r = 1

n2

n!

r= limn

|an+1||an| = limn

(n+ 1)2

n2n!

(n+ 1)! = lim

n(n+ 1)2

n2(n+ 1) = lim

nn+ 1

n2 =

LHopital= lim

n1

2n=

1

= 0< 1

nn

n=1

n

n 12

n

limn

n

|an| = limn

n

n 12

= 1 12

=1

2


20/35

an(x c)n

R

|x

c|

< R

R

c

|x c| > R R

(x c)

n

sen x=

n=0

(1)nx2n+1(2n+ 1)!

2n+ 1

00 = 1

0! = 1

R

(x c)

x

nen

x3n+5

an = nen

x3n+5

x c

r = limn

|an+1||an| = limn

(n+ 1)

n

en

en+1|x|3(n+1)+5

|x|3n+5 = limn(n+ 1)

ne |x|3 =

LHopital= lim

n|x|3

e =

|x|3e

.

n

n

r


21/35

r= limn

an+1(x 1)n+1+12 an(x 1)n+12 = limn(n+ 1)

n

2n+1

2n |x 1| 12 = lim

n(n+ 1)2

n |x 1| 12 =

r = lim

n2

1|x 1| 12 = 2|x 1| 12

r= 2

|x

1

|12


22/35

xn

n

r= limn

|an+1xn+1||anxn| = limn

(n+ 1)

n

|x|n+1|x|n = limn

(n+ 1)|x|n

=

LHopital= lim

n|x|1

= |x|

r


23/35

ln(1 + x) =

n=0

(1)nxn+1n+ 1

1< x 1 ln x

anx

n

ln 0 =

x= 0

x

f(x) = ln(1 +x)

x= 0

f(x) = (ln(1 +x)) = 1

1+x

n=0 aorn = a0

1 r

f(x) = 1

1 +x=

n=0

(x)n =n=0

(1)nxn

r= x

|r| = | x| = |x|


24/35

f(x) =

n=0an(x c)n an= f

(n)(c)n!

f

(x c)0

f(k)(x) =n=k

ann (n 1) (n k+ 1)(x c)nk

f(k)(c) = akk (k1) (k k + 1) = akk!

ak = f(k)(c)

k!

f

N+ 1

c

x

f(x) =

Nn=0

f(n)(c)(x c)nn!

+RN Rn = f(n+1)(z)(xc)n+1

(n+1)!

z [c, x]

f(x) =pN(x) +RN pN(x) =a0+a1(x c) + +aN(x c)N

N f

ai=

f(i)(c)i!

pn(x) =

Nn=0

f(n)(c)(x c)nn!

N

c

f(x)

|Rn| Mn+1|xc|

n+1

(n+1)! Mn+1

|f(n+1(z)| |f(n+)(z)| Mn+1 z

x

n

n=0

f(n)(c)(x c)nn!

c

c= 0

x

c

x


25/35

sen0.1

3

sen x = cos x

sen x = sen x

sen x = cos x

sen(4) x = sen x

0.1

0

0

p3(x) =f(0) +f(0)(x 0) +f(0) (x0)2

2! +f(0) (x0)

3

3!

f(0) = sen 0 = 0 f(0) = cos 0 = 1

f(0) =

sen 0 = 0

f(0) =

cos0 =

1

p3(x) = 0 +x +

0x2

2! +x

3

3! = x

x3

6

sen0.1 0.1 0.13

6 = 0.1 0.0016 = 0.1 0.00016666 = 0.0998333

M4 max{|f(4)(z)|} = max{| sen z|} z [c, x] = [0, 0.1]

| sen | 1

M4= 1 |R3| M4 |x|4

4! = 0.1

4

24 = 0.0001

24

lim

NRN= 0 f(x) =

n=0

f(n)(c)(xc)nn!

f

|f(n+1)(z)| M

n

z [c, x]

M

n

z

C

f(x) =

e

1x2 , x = 0

0 , x= 0

f(k)(0) = 0

k

0

n=0

0xn

n! =

n=0

0 = 0

x

f(x)

x = 0

f(x)

0

ex =

n=0xn

n!

x.

f(x) =ex

f(x) =ex

f(x) =ex

f(n)(x) =ex

c= 0

0

f(0) =e0 = 1

f(n)(0) =e0 = 1

n

0

n=0

f(n)(c)(x c)nn!

=n=0

f(n)(0)xn

n! =

n=0

xn

n!

f(x)

n

|f(n+1)(z)| =|ez| = ez

[c, x] = [0, x]

M

[0, x]

M |f(n+1)(z)| M z [c, x] = [0, x] |f(n+1)(z)| =

|ez| = ez n M n fn+1(z) n Mn

n

|f(n+1)(z)|

M

n

lim

N|RN| = lim

N

f(N+1)(zN)xN(N+ 1)!

limN

MxN

N! =M 0 = 0

f(x) =

n=0

xn

n!

sen x=

n=0(1)nx2n+1

(2n+ 1)!

x.

f(x) = sen x

f(x) =

cos x f(x) = sen x

f(3)(x) = cos x

f(4)(x) = sen x = f(x)

0


26/35

| sen x| 1 | cos x| 1

|fn+1(z)| 1 = M

n

limN

|RN| = limN

f(N+1)(zN) xN(N+ 1)!

M limN

xN

N! =M 0 = 0,

sen x

x

c = 0

f(0) = sen0 = 0

f(0) = cos0 = 1

f(0) = sen 0 = 0f(3)(0) = cos 0 = 1

f

1

n= 2k+ 1

f(n)(0) =f2k+1(0) = (1)k 0

n=0

f(n)(0)(x 0)nn!

=k=0

f(2k+1)(0)x2k=1

(2k+ 1)! =

k=0

(1)kx2k+1(2k+ 1)!

n

ln x=

n=0

(1)n(x 1)n+1n+ 1

12 x 2

f(x) = ln x

f(x) = 1

x f(x) =1

x2 f(x) = 12

x3 f(4)(x) =123

x4

f(5)(x) = 1234x5

, . . .

f(n)(x) = (1)n+1(n1)!xn

n >0

f(n)(1) = (1)

n+1(n1)!1n+1

= (1)n+1(n 1)!

n >0

f(x) =n=0

f(n)(1)(x 1)nn!

=f(1) +n=1

f(n)(1)(x 1)nn!

= 0 +n=1

(1)n+1(n 1)!(x 1)nn!

=n=1

(1)n+1(x 1)nn

=k=0

(1)k(x 1)kk+ 1

.

ln x

|f(n+1)(z)| =

n!zn+1

z

1 z x

z= 1

f(n+1)(z) n!

limN

|Rn| = limn

f(n+1)(zn) |x 1|n+1(n+ 1)!

limN

Mn|x 1|n+1(n+ 1)!

= limn

n!|x 1|n+1(n+ 1)!

= limn

|x 1|n+1n+ 1

= 0

|x 1| 1 [1, 2]

12

, 1

f(n+1)(z)

x

x z 1

|f(n+1)(z)| =

n!zn

n!

xn+1 =Mn+1 0< x


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12 x 1

x1x

1

0, 1

2

ln x

x

]0, 2]

cos x=

n=0(1)nx2n

(2n)!

x

cos x

2

sen(x2)

0

sen x


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x

{1, x , x2, x3, . . .}

x

{1, cos x, sen x, cos(2x), sen(2x), cos(3x), sen(3x), . . .}

a02

+n=1

(ancos(nx) +bnsen(nx))

ak bk

f(x) =a0+

n=1

(ancos(nx) +bnsen(nx)) [, ]

a0 = 1

f(x)dx

k >0

ak =

1

f(x)cos(kx)dx

bk= 1

f(x)sen(kx)dx

f(x) = a02

+n=1

(ancos(nx) +bnsen(nx))

[, ]

[, ]


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f(x)dx=

a02

dx+n=1

an

cos(nx)dx+bn

sen(nx)dx

cos(nx)dx=

sen(nx)dx= 0

f(x)dx= a0 a0 =

1

f(x)dx

ak bk k = 1, 2, 3, . . . f(x)cos(kx)

f(x) sen(kx)

[, ] cos2(kx)dx= sen2(kx)dx= cos(nx)sen(kx)dx= 0 n k n=k

cos(nx)cos(kx)dx=

sen(nx)sen(kx)dx= 0

f(x) cos(kx)dx= a0

cos(nx)dx+n=1

an

cos(nx)cos(kx)dx+bn

sen(nx)cos(kx)dx

f(x) cos(kx)dx= ak

ak = 1

f(x)cos(kx)dx

f(x) sen(kx)dx= bk

bk= 1

f(x)sen(kx)dx

f : R R 2 f(x+ 2) =f(x) x

x

f

x

f(x+)+f(x)2

f

x

f

x

f(x)

f

f(x) = f(x)

x

f(x) = f(x) x

f

LLf(x)dx = 2

L0

f(x)dx

f L

Lf(x)dx= 0

f(x) = xn

n

n

cos(nx)

sen(nx)


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f(x) =

|x|x

, x = 00 x= 0

[, ]

a0 = 0 f(x)cos(nx) ak = 0

k= 1, 2, . . .

bk f(x)sen(nx)

bn = 1

|x|x

sen(nx)dx= 2

0

sen(nx)dx=2

cos(nx)n

0

= 2 cos(n)+2cos0n

=

0 , n 4

n , n

f(x) =k=0

4

(2k+ 1)sen ((2k+ 1)x)

x =

2

1 = 4

k=0

1

2k+ 1sen

(2k+ 1)

2

sen

(2k+ 1) 2

=

1 , k 1 , k 1 = 4 k=0

(1)k2k+ 1

4 =

k=0

(1)k2k+ 1

f(x) = |x| [, ]

f(x)sen(nx)

bk = 0 k= 1, 2, . . .

ak f(x)cos(nx)

a0 = 1

f(x)dx= 2

0 xdx=

an = 1 |x| cos(nx)dx = 2 0 x cos(nx)dx = 2x sen(nx)n cos(nx)n2

0= 2

n2 (cos(n) cos0) =

0 , n 4n2

, n

f(x) =

2+

k=0

4(2k+ 1)2

cos ((2k+ 1)x) =

2 4

k=0

cos((2k+1)x)(2k+1)2

f(x) =x

[, [

f [a, b] f(x+kT) =f(x)

k

T = b a

f

T

f

[a, b]

f

f

f

[a, b]

f

L= T

2 = ba

2

ao = 1

L

LL

f(t)dt= 1

L

ba

f(t)dt

n= 1, 2, 3, . . .

an = 1

L L

Lf(t)cos

n

L

t dt= 1

L b

a

f(t)cosn

L

t dtbn =

1

L

LL

f(t)senn

Lt

dt= 1

L

ba

f(t)senn

Lt

dt


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a02

+n=1

ancos

nL

t

+bnsenn

Lt

f(x) = |x|

[, ]

x=

2

8 =

k=0

1

(2k+ 1)2

f(x) =x

[0, 1[

f

x= 2

f(x) = 1

[1, 1]

g(x) = |x|

[1.1[

h(x) = 1 |2x|

[1, 1[


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M

n, |xn| M

M

X R

x X, x M X R

X R sup X= inf{M R :M X}

S= sup X

>0

x X :S < x

x S x S X

S

S

S= sup{xn} lim

nxn=S

>0

S

{xn} N S < xN< S

xn

xN n > N S

< xN < xn S

xn S S < xn S < S+ N : n > N = S < xn < S+

limn

xn = S


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xn+1=(xn)

a

a

a


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n=0

(1)nn+ 1

S=

n=0

(1)nn+ 1

= 1 12

+1

31

4+

1

2S=

1

2

n=0

(1)nn+ 1

= 1

2 1 1

2 2+ 1

2 3 1

2 4+

= 0 +

1

2 1+ 0 1

2 2+ 0 + 1

2 3+ 0 1

2 4+

S

12

S

3

2S= 1 + 0 +

1

3 2

2 2+1

5+ 0 +

1

7 2

2 4+ 0 +

= 1 +1

31

2+

1

5+

1

71

4+ =

n=0

1

4n 3+ 1

4n 1 1

2n

.

3

2

n=0

(1)nn+ 1

=n=0

14n 3+

1

4n 1 1

2n

32

S=S

n=0

1

4n 3+ 1

4n 1 1

2n

= 1 +

1

31

2+

1

5+

1

71

4+

(1)nn+ 1

= 1 12

+1

31

4+


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Sequencias e Series

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Transcript of Sequencias e Series