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 +…

SOLUTION: City Colleges of Chicago Calculus Questions

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