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 +…