5 Steps to a 5 AP Calculus BC 2019

Series 367

x/=

√

5

7

. Therefore,f(x) is a strictly
decreasing function forx≥1 and

x/=

√

5

7

. Thus, for the series
∑∞
n= 1

(−1)n

n+ 1

7 n^2 − 5

,an≥an+ 1. Also,

note that limn→∞

n+ 1

7 n^2 − 5

nlim→∞

1

14 n

= 0

(by usingL’Hoˆpital’sRule).

Therefore, the alternating

series

∑∞

n= 1

(−1)n

n+ 1

7 n^2 − 5

converges, and since

the series does not converge absolutely as

shown earlier, it converges conditionally.

9. The sum of the geometric series
   ∑∞
   n= 0

4

(

1

3

)n

isS=

an

1 −r

=

4

1 − 1 / 3

= 6.

10. For the alternating series

∑∞

n= 1

(−1)n−^1

2 n− 1

approximated bys 50 , the maximum

absolute error|Rn|<an+ 1 ,so

|R 50 |<a 51 =

(−1)^50

101

=

1

101

≈ 0 .0099.

11. Examine the ratio of successive terms.∣
    ∣∣
    ∣

xn+^1

1 +(n+1)^2

·

1 +n^2

xn

∣∣

∣∣=

∣∣

∣∣ x(1+n

(^2) )
n^2 + 2 n+ 2
∣∣
∣∣.
Since limn→∞
∣∣
∣∣ x(1+n
(^2) )
n^2 + 2 n+ 2
∣∣
∣∣=|x|, the series
will converge when|x|<1or
− 1 <x<1. Whenx=1, the series
becomes
∑∞
n= 0

1

1 +n^2

. This series is
term-by-term smaller than thep-series
withp=2; therefore, the series converges.
Whenx=−1, the series becomes
∑∞
n= 0

(−1)n

1 +n^2

, which also converges.

Therefore, the interval of convergence

is [−1, 1].

12. The ratio is

∣∣

∣∣(3x)

n+ 1

(n+1)^2

·

n^2

(3x)n

∣∣

∣∣=

∣∣

∣∣^3 xn

2

(n+1)^2

∣∣

∣∣

and limn→∞

∣∣

∣

∣

3 xn^2

(n+1)^2

∣∣

∣

∣=|^3 x|so the series will

converge when| 3 x|<1. This tells you

that− 1 < 3 x<1 and

− 1

3

<x<

1

3

.

Whenx=

1

3

, the series becomes

∑∞

n= 1

3 n

n^2

(

1

3

)n

=

∑∞

n= 1

1

n^2

, which is a

convergentp-series. Whenx=−

1

3

, the

series becomes

∑∞

n= 1

3 n

n^2

(

−

1

3

)n

=

∑∞

n= 1

(−1)n

n^2

,

a convergent alternating series. Therefore,

the interval of convergence is

[

−

1

3

,

1

3

]

.

13. Represent lnxby a Taylor series.
    Investigate the first few terms by finding
    and evaluating the derivatives and
    generating the first few terms.
    f(a)=lna,f′(a)=

1

a

,f′′(a)=

− 1

a^2

,

f′′′(a)=

2

a^3

,solnxcan be represented by

the series=

lna

0!

(x−a)^0 +

1 /a

1!

(x−a)^1

+

− 1 /a^2

2!

(x−a)^2 +

2 /a^3

3!

(x−a)^3 +···

=lna+

(x−a)

a

−

(x−a)^2

2 a^2

+

(x−a)^3

3 a^3

+···+

(−1)n−^1 (x−a)n

nan

+···

Using the ratio test,

nlim→∞

∣∣

∣

∣

(−1)n(x−a)n+^1

(n+1)an+^1

·

nan

(−1)n−^1 (x−a)n

∣∣

∣

∣=

nlim→∞

∣∣

∣∣ n

(n+1)

·

(x−a)

a

∣∣

∣∣=

∣

∣∣x−a

a

∣

∣∣.

The series converges when

∣

∣∣x−a

a

∣

∣∣<1,

that is,− 1 <

x−a

a

<1. Solving the

inequality, you find−a<x−a<aor

0 <x< 2 a. Whenx=0, the series becomes

∑∞

n= 1

(−1)n−^1 (−a)n

nan

=

∑∞

n= 1

(−1)^2 n−^1

n

=

∑∞

n= 1

− 1

n

=−

∑∞

n= 1

1

n

.