Review Test 3: 6 Multiple Choice: Series: Convergence, Divergence, Absolute Convergence, Conditional Convergence, Sum (geometric, telescoping) Free Response: 1. L’Hopital – recognize and apply 2. Improper Integrals – recognize type of improper integral, compute improper integrals using CORRECT notation 3. Series – use known tests (alternating series, root, ratio, p-series, limit comparison, integral, basic comparison, geometric, basic divergence) to determine convergence 4. Taylor Polynomials and Series – Give Taylor polynomials using given information (values, functions, etc); be able to find the error; radius and interval of convergence

Find a Taylor series for cos (x2) centered at x = 0: Find a Taylor series for e2x centered at x = 0:

Find the Taylor polynomial P5(x) for f (x) = xcosx2. Find the nth Taylor polynomial Pn for the function f (x) = e–x Find the nth Taylor polynomial Pn for the function f (x) = sinh x

Find the nth Taylor polynomial Pn for the function f (x) = ln (1 – x) Give the 5th degree Taylor polynomial for f (x) = sin(x) centered at 0. Give the 5th degree Taylor polynomial for f (x) = ex centered at 0.

Give the 5th degree Taylor polynomial for f (x) = ln(x+1) centered at 0. Give the 5th degree Taylor polynomial for f (x) = cos(x) centered at 0.

( ) ( ) ( )2 1 2 2 2 1f , f ' , f ''= − = = − Give the 2nd degree Taylor polynomial

for f centered at 2. Rewrite f (x) = 3x3 +2x2 – x + 1 in powers of (x – 2).

Create the 3rd degree Taylor Polynomial for f (x) = arctan(x) centered at x = 0.

( ) ( )( )n 1

n 1n

cR x x

n 1

f

!

++=

+

Use the Lagrange formula to find the smallest value of n so that the Taylor polynomial of degree n for f (x) = cos (x) centered at x = 0 can be used to approximate f (x) within 10 –4 at x = 1.

Use the Lagrange formula to find the smallest value of n so that the nth degree Taylor Polynomial for f (x) = ln (1 + x) centered at x = 0 approximates ln (2) with an error of no more than 0.01.  

Which term is truncated if we want to approximate the sum of ( )n 13

n 1

12n 1

+∞

=

−−∑

with an error of less than 11000

?

 

1. State the indeterminate form and compute the following limits :

a. ( )

n

n 4

n 2

lnl im→∞

+

+

b. ( )2n

n3nl im

→∞

c. 2n

n

31n

l im→∞

⎛ ⎞+⎜ ⎟⎝ ⎠

d. ( )( )x 0

x 2x

x 2x

sinl im

sin→

−

+

e. 2x

2x 0

e 12x

l im→

−

f. x

x 0

1x

l im+→

⎛ ⎞⎜ ⎟⎝ ⎠

g. ( )x 3

2x 0

3e 3 x

x

/

l im→

− +

h. 2

x

xx

l imln→∞

i. ( )

x

xx 0

1 x e

x e 1l im→

+ −

−

j. ( )

x 0

4x

x

arctanl im→

 

2. Give the exact value of 0

12nn

∞

=∑ .

3. Give the exact value of ( )2

11n n n

∞

= +∑ .

4. Give the exact value of ( )

2 3nn

ncos∞

=

π∑ .

5. Evaluate each improper integral, and explain why it is improper. Use correct notation.

a. 2

21

1dx

x−∫

b. 1

0 6

1dx

1 x−∫

c. ( )

7

25

14dx

x 6−∫

d. 27

2 3

0

x dx/−∫

e.

4

0

1dx

4 x−∫

f. 20

1dx

1 x

∞=

+∫

g. 5

2

dxx 2

=−∫

Notes for series “growth”: Let p(k) be a polynomial in k. rk for r > 1 grows much faster than p(k) k! grows much faster than rk, p(k) kk grows much faster than the others Hence,

( ) ( ) ( )k k

k k

k k

p k p k p k

kr k

r r kk k k

, ,!

!, ,!

∑ ∑ ∑

∑ ∑ ∑

ALL converge rapidly.

Determine if the following series converge absolutely, converge conditionally, or diverge?

a. ( ) 1

1

13

n

n

nn

+∞

=

−+∑

b. 21

cosn

nn

∞

=∑ π

c. ( )n2

n 0

4n 1

3n 2n 1

∞

=

−

+ +∑

d. ( )n2n 0

3 1

3n 2n 1

∞

=

−

+ +∑

e. ( )n2n 0

3n 1

3n 2n 1

∞

=

−

+ +∑

f. ( )n

n

n 0

n4 1

n 3

∞

=

⎛ ⎞⎛ ⎞⎜ ⎟− ⎜ ⎟⎜ ⎟+⎝ ⎠⎝ ⎠∑

g. ( )n

2 3n 0

2 1 n

3 n n

arctan∞

=

⎛ ⎞−⎜ ⎟⎜ ⎟+ +⎝ ⎠

∑

h. ( )n n

nn 0

1 3

4 3n

∞

=

⎛ ⎞−⎜ ⎟⎜ ⎟+⎝ ⎠

∑

i. ( )

( ) ( )

n

n 0

1 3

n 2 n 2ln

∞

=

⎛ ⎞−⎜ ⎟⎜ ⎟+ +⎝ ⎠

∑

j. ( )( )

n

n 2

1 n

n 1

!

!

∞

=

−

+∑

k. ( )n

n 2

1

3n 2

∞

=

−

+∑

l. ( )n 2

nn 2

1 10n

3

∞

=

−∑

m. ( )n n

n 2

1 3

n!

∞

=

−∑

n. ( )n2

n 2

1

n 3n 2

∞

=

−

+ +∑

o. ( ) n

n 2

n n

n

cos!

∞

=

π∑

p. ( )( )2n 2

1

n nln

∞

=∑

Converge or diverge? Additional review problems.

a. 2

5n 2

n 3n 24n n 1

∞

=

+ −+ −∑

b. 2

6n 1

n 3n 2

4n n 1

∞

=

+ −

+ −∑

c. 5

nn 1

n5

∞

=∑

d. ( )1

1 .1n n n

∞

= +∑

e. 31

1

n n

∞

=∑

f. 31 2n

n

n n

∞

= +∑

g. 0

27nn

∞

=∑

h. ( )2

1

1 n

n n

∞

=

−∑

i. 1

1 11n n n

∞

=

⎛ ⎞−⎜ ⎟+⎝ ⎠∑

j. 1

52 1n n

∞

= −∑

k.

2

1

3!

n

n n

∞

=∑

l. 1

25 1

n

n

nn

∞

=

⎛ ⎞⎜ ⎟−⎝ ⎠

∑

m. 1 2

31

( 1)3 1

n

n

nn

−∞

=

−+∑

n. 0

532

n

n

∞

=

⎛ ⎞−⎜ ⎟⎝ ⎠∑

o. 1n

nn

∞

=∑

p. 1

11 n

n e

∞

−= +∑

q. 31

5n

n n

∞

=∑

r. 1cos( )

nn

∞

=∑ π

s. ( )2n 2

1

n nln

∞

=∑

t.

3n

n 1

ne∞

−

=∑

u.

n

n 1

nn 1

∞

=

⎛ ⎞⎜ ⎟+⎝ ⎠

∑

v. 31

11n n

∞

= +∑

w. 1

!n

n

ne

∞

=∑

x. ( )n

nn 2

1 n

n

!∞

=

−∑

y. ( )n

n 2

1 n

3n 2

!∞

=

−

+∑

z. ( )( )

n

n 2

1 n

n n 1

!

!

∞

=

−

+∑

aa. ( ) ( )n

2n 2

1 n 1

5n 2n 1

∞

=

− −

+ −∑

bb. ( )

n 2

n

n 7

cos∞

=

π

+∑

cc.( )n n

nn 2

1 2

2 1

∞

=

−

+∑

dd. ( )2

n 2

n

1 n

arctan∞

= +∑