# SOLUTION: City Colleges of Chicago Calculus Questions

SOLUTION: City Colleges of Chicago Calculus Questions.

View attached explanation and answer. Let me know if you have any questions.

(1) f ( x ) = n5n x n

n =1

an +1

an

( a ) L = lim

n →

= lim

; an = n 5 n x n

( n + 1) 5n+1 x n+1

n5n x n

n →

1

= lim 5 x 1 +

n →

n

= 5x

L 1 5x 1 x

1

5

Radius of convergence = R =

1

5

( b ) f ( x ) = n5n x n = (1) 51 x1 + ( 2 ) 52 x 2 + ( 3) 53 x3 + ( 4 ) 54 x 4 + …

n =1

f ‘ ( x ) = (1) 51 (1) + ( 2 ) 52 ( 2 ) x + ( 3) 53 ( 3) x 2 + ( 4 ) 54 ( 4 ) x3 + …

f ” ( x ) = 0 + ( 2 ) 52 ( 2 )(1) + ( 3) 53 ( 3)( 2 ) x + ( 4 ) 54 ( 4 )( 3 ) x 2 + …

f ” ( 0 ) = ( 2 ) 52 ( 2 )(1) + 0 + 0 + …

f ” ( 0 ) = 100

( c ) At x = −

1

,

5

( −1)

n

1

n5 x = n5 − = n5n n = ( −1) n, which is divergent series by divergence test.

5

5

n =1

n =1

n =1

n =1

1

So, the series diverges at x = − .

5

n

n

n

n

n

( 2) ( a )

2n

2n

,

a

=

n

3

n3

n =1 n

an +1

2n +1

n3

= lim

3

n

n → a

n →

( n + 1) 2

n

lim

= 2 lim

n →

n3

( n + 1)

3

1

= 2 lim

n →

=2

1

1 +

n

1

(1 + 0 )

3

3

= 2 1

2n

diverges.

3

n =1 n

So, by Ratio Test, the series

(b)

( −1)

n

ln n

n=2

, an =

1

ln n

an +1 an and liman = lim

n →

n →

1

=0

ln n

So, by Alternating series test, the series

n=2

(c)

n =5

an =

1

, an =

( −1)

n

ln n

converges.

1

n −2

n −2

1

1

1

1

; bn =

= 1/ 2

n −2

n

n n

an bn

Since

bn is a diverging p-series and an bn , so by Comparison Test, the series

n =5

n =5

1

n −2

diverges.

( 3) The region enclosed by the curves y = 0, y = x 2 + x between x=0 and x=1 along with

axis of revolution are shown below:

We will use washer method to find the volume.

Volume = ( r2 2 − r12 ) dx

b

a

1

2

2

= ( x 2 + x + 1) − ( 0 + 1) dx

0

1

= x 4 + x 2 + 1 + 2 x3 + 2 x 2 + 2 x − 1 dx

0

1

= x 4 + 2 x3 + 3x 2 + 2 x dx

0

27

=

10

27

Hence, the required volume is

.

10

( 4) (a )

( −2 ) ( x + 4 )

n

n+3

n =1

L = lim

n →

( −2 ) ( x + 4 )

n

, an =

n

n+3

an +1

an

( −2 ) ( x + 4 )

n +1

= lim

n

n +1

n+4

n →

= lim −2 ( x + 4 )

n →

n+3

( −2 ) ( x + 4 )

n

n

n+3

n+4

3

n

= lim −2 ( x + 4 )

n →

4

1+

n

1+ 0

= −2 ( x + 4 )

1+ 0

1+

=2 x+4

L 1 2 x + 4 1 x + 4

9

At x = − ;

2

n =1

( −2 ) ( x + 4 )

n

1

1

1

9

7

− x+4 − x−

2

2

2

2

2

n

1

n

9

−2 )

(

−

2

−

+

4

n

(

)

( −2 )

1

2

=

=

=

n+3

n+3

n =1

n =1

n =1 n +…