Answers to Exam 3 Review
1) 0, 2, 15, 679
2) a. (i) 6, 7;
Math 401—Calculus II
(ii) a1 = 1; an+1 = (1)n(|an| + 1)
b. (i) 36, 49
(ii) a1 = 1; an+1 =

(iii) an = (1)n+1 n

an  1
2
(iii) an = n2
3) a.  1, 21 , 31 , 41 ,... which converges to 0
b. 100 ,100 ,100 ,100 ,... which obviously converges to 100
4,4.9,4.99,4.999,... which converges to 5
4)
(Note: 4.9 equals 5 exactly)
5) a. S1 = 31 , S2 = 52 , S3 = 73 , S4 = 49
b. S n  2 nn1
c. The partial sums converge to ½, which is the value of the series.
6) a.
b. Except for a finite number of terms, this sequence is just an  ne  n , so it has the
same limit. Note that lim enn  lim e1n  0 , by L’Hopital’s Rule.
3
2
n
n
n
c. Since n grows more quickly than n!, the limit is zero. This can be proven by induction , but we
haven’t learned proofs by induction, so sequences with n! are the only ones that I will allow
referencing our hierarchy of growth rates for. One approach to mathematically showing this
sequence goes to zero, though, would be this: Consider the infinite series defined by summing the
n
terms of this sequence. Using the Ratio Test, the ratio of two successive terms gives  nn1  , whose
limit is 1e , so the infinite series converges, which can only happen if the terms of the sequence
approach zero. Pretty slick.
d. The even terms converge to 1, while the odd terms converge to  1. Thus, the sequence as a whole
does not converge.
e. Since  1  sin n  1 for all n, the given sequence satisfies 2n1  sin2nn  2n1 , and since both ends go
to 0, the given sequence converges to zero as well, by the Squeeze Theorem.
7) a. Since ln n12  12 ln n , we can use the growth rates of {ln n} and {np} or L’Hopital’s Rule to show
 
 .
that 12 ln n n
b. Since
1
2

1
10
1
2
 n .
, our growth-rates comparison shows that n
1   43 
b. S  1 
1  43
c. e
d.  52
1
1
10
2
10
8) a. S  16 
9) a.
129
12
 8176
5
2
b.
64
49

1
10) 1.00393939… = 1   .0039100
 =1+
11) a. S n 
12) a.
113
30

1 1
3 1
k
k 0

1
3 n 4
, S =
b.
1
3
39
 1  9900

b. S n 
1
2

1
3

9939
9900

1
n13
3313
3300

1
n 3
2
5
13) a. lim an  0 , so the test is inconclusive.
n
c. lim an  1 , so it diverges.
n
.0039
.99
b. lim an  1 , so it diverges.
n
14) a. Let f ( x ) 
1
x ln x
, which is continuous and decreasing on (1,). Since

 f ( x )dx   , the series
1
diverges.
b. Let f ( x ) 
x
x 2 4
. Then f(x) is continuous for x  1. However, f’(x) =
4
3
 x 2  4 


 0 , so f is
increasing, and the conditions of the Integral Test are not satisfied. The Divergence Test, though,
will show the series diverges.
14. c. Let f ( x ) 
, which is continuous and decreasing on (1,). Since
1
x ln x 2

 f ( x )dx 
1
ln 2
, a finite
1
number, the original series converges.

15. a.

 
k 2

b.
1
k
e
is a p-Series with p =   e < 1, so the series diverges.

4
4
  k 12     k1 
k 3
by a change of index, so this is a p-Series with p = 4 (p > 1), so it converges.
k 1
16. a. The ratio between successive terms ends up being
2
k 1
 k k1 k is the reciprocal of  kk1 k  1  k1 k = e.
k k
2
1
k 1   k 1  is 0 e = 0, so the series converges.
limit of
  k k1  . The limit of the first part is 0. The
k
Thus, lim  k k1   1e . Therefore, the limit of
k
k 
b. The ratio between successive terms ends up being k 11   k k1  , and the limit as k  is 0, so the
series converges.
2
c. The ratio between successive terms ends up being 2 k k212 k 1 , and the limit as k  is 41 , so the
series converges.
17. a. lim k ak  lim k2k1  21  1 , so the series converges.
6
k 
k 
b. lim k ak  lim k1  0  1
k 
18. a.
b.
k 

1
k 2 4
1
k
3
2 1

1
k2
1
k
3
and
2
1
k2
converges, so the original converges by the Comparison Test.
1
and
k
3
2
is a p-series with p > 1 and thus converges, so the original converges by the
Comparison Test.
c. Use the Limit Comparison Test with
 , which converges.
1
3k
The ratio of the terms of the two
series has a limit of 1 as k. Since the comparison converges, the given series does as well.
d. k 1k 2  k 1 k  13 2 . Since the terms of 13 2 are a p-series with p > 1, it converges, so the original
k
k
does as well.
k
19. a. Use the Ratio Test. The ratio between successive terms ends up being 2   k k1  , which has limit
2
e
< 1. Therefore, the series converges.
b. ln kk21   lnk  2  lnk  1 , so the series telescopes. We get Sn = ln(n + 2)  ln 2, and the limit
as n is . Since the sequence of partial sums diverges, the series diverges.
  

c.
k 1
k 1
k2

k 1
1
k

 k12 , and since
   converges (p-series, p >1), if the original series also

k 1
1
k2

converged, then the sum of them would converge, but the sum of the series would be
   , which
k 1
we know to be divergent (Harmonic Series). Thus, the original series cannot converge.
1
k