BC Calc III
Quiz 9.4 - 9.5
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#1 (4 pts) Find the interval of convergence of (No work necessary).

a.
b.
c.
d.
xn

n
n 0 k  4

(x  3)n

3n
n 0

( x  2) n

n2
n 0

(1)n ( x  3)n

n  5n
n2
Interval of convergence = [4, 4)
Interval of convergence = (0, 6)
Interval of convergence = [ 3, 1]
Interval of convergence = (8, 2]
#2 (5 pts) Find the interval of convergence of (Show all work).
2n ( x  2) n

n2
n 0

Absolute Ratio Test:
2k2  x  2 
 ( x  2) k 1  2k 1

k2
lim 

  2 x  2  L
  lim 
k 
(k  1) 2
( x  2) k  2k  k   (k  1) 2 

5
3
So if 2  x  2  1, then   x  
2
2
n
n


5
2 ( x  2)
(1)n
Endpoints: x    
which converges by A.S.T.


2
2
n2
n 0
n 1 n

3
2n ( x  2)n  1
x 
  2 , which converges by p-test.
2 n 0
n2
k 1 k
Therefore the interval of convergence is
BC CALC III
 5 3
  2 ,  2 
(1)n
.
n  3n
n 1
a. Show that this series converges.
1
1
 1 
Since

 0 and lim 
 0,
n
n 1
n 
n 
n 3
n 1  3
 n 3 

#3(8 pts). Look at the series



n 1
(1)n
converges by
n  3n
AST.
b. Find a value for n such that Sn is within .001 of the actual sum.
1
S  Sn  an 1 
n  1  3n 1
1
1
Since

 .001 ,
51
5 1  3
729  5  1
We may take n = 5.
#4(5 pts). Determine whether the following series converges conditionally, converges
absolutely, or diverges? Show all steps/explain.

(1)n  n ! nn

(2n)!
n 1
Look for Absolute Convergence using the ratio test:


k 1

(1) n  n ! n n
n ! n n

.
(2n)!
k 1 (2n)!
n

 (n  1)! (n  1) n 1 (2n)! 
(n  1) 2
 n 1   e
Then, lim 

 lim 

   1
n 
(2n  2)!
n ! n n  n  (2n  2)(2n  1)  n   4


Therefore,

n 1
BC CALC III
(1)n  n ! nn
converges absolutely.
(2n)!

#5(4 pts) Suppose that the power series
 a ( x  1)
n0
n
n
converges if x  3 and diverges if x = 8.
Indicate whether the following statements must be true, which may be true, and which cannot be
true. Justify your answers.
a. The power series converges if x = 2.
Must be true. Since the series converges at x  3 , which is 4 units from the
center at x = 1, the radius of convergence is at least 4. Since the distance from
x = 2 to x = 1, is 1 < 4 <R. The series must converge at x = 2.
b. The power series converges if x  4 .
May be true. Since 4  R  7 and 4 1  5 , the series may converge or diverge
at x  4
c. The power series diverges if x = 5.
May be true. If R = 4, the x = 5 would be an endpoint of the interval opf
convergence and the series may converge or diverge at an endpoint.
d. The power series converges absolutely if x = 4.
Must be true. Since x = 4 is only 3 units from the center, it is necessarily in the
interior of the interval of convergence and a power series convereges absolutely
on the interior of the interval of convergence.
1,000,000
1
to the nearest whole number. Explain all analysis clearly.
n
n 1
1,000,000
1,000,000
1
ln(n) dn  
 a1  1, 000, 000
By the Integral Test, we know 
1
n
n 1
#6 (2 pts) Estimate
BC CALC III
