Calculus Sequences and Series Solutions
7
FAMAT State Convention 2015
7
7
( a1 + a7 ) = ( a1 + a1 + 6d) = 7a1 + 21d = 119 . Therefore a1 + 3d = 17 . Since all the
2
2
n=1
terms of the sequence are positive integers, a1 and d are both positive integers. Therefore
the largest value of d will occur when a1 = 2 and hence d = 5 . B
1.
åa
n
=
( )
2. a3 = 3 and a9 = a3r 6 = 6 , so r 6 = 2 . Therefore a27 = a3r 24 = a3 r 6
4
= 3× 2 4 = 48. C
¥
¥ 1¥
¥ 1¥
3. lim ¥1+ ¥ = e¥ 0 , therefore ¥ ¥1+ ¥ diverges by the nth Term Test for Divergence.
n®¥ ¥
¥ n¥
n¥
n=1
C
2
¥
x2
n n
4. å ( -1) 3
converges by the Alternating Series Test since f ( x) = 3
is a
x +1
n +1
n=0
¥
n2
n2
= 0 . However, å 3
decreasing function (for x ³ 2 ) and lim 3
diverges by
n®¥ n +1
n=0 n +1
comparison to the divergent Harmonic Series by the Limit Comparison Test. Therefore
2
¥
n n
-1
( ) 3 converges conditionally. B
å
n +1
n=0
n
n
n
-1) 2 n n!
(
5. Let an =
. Then
2 × 5×8 (3n -1)
2 n+1 ( n +1)!
2 ¥5¥8 (3n -1)
2 ( n +1)
a
lim n+1 = lim
¥
= lim
n
n®¥ a
n®¥ 2 ¥5¥8
n®¥ 3n + 2
2 n!
(3n -1) ¥3n
( + 2)
n
¥
( -1)n 2n n! converges absolutely by the Ratio Test. A
å
(3n -1)
n=1 2 × 5×8
2 + cosn 1
³ for all n and hence
n
n
¥
1
to the divergent Harmonic Series å . C
n=1 n
6. Notice that
=
2
< 1. Therefore
3
2 + cosn
diverges by direct comparison
n
n=1
¥
å
1/n
¥¥1 1 ¥n ¥
¥1 1 ¥ 1
7. lim ¥¥ + ¥ ¥ = lim ¥ + ¥ = < 1. Therefore
n®¥ ¥2
n®¥ ¥2
n¥ ¥
n¥ 2
¥
¥
n
¥1 1 ¥
+ ¥ converges absolutely by
¥
¥
¥
n¥
n=1 2
the Root Test. A
2n
+ ln 2
2n
8. Let lim an = L . Then ln L = lim
=
lim
= ln 2 , where
n®¥
n®¥
n®¥
n®¥
n2
n
n2 + 2 n
+1
2n
the second equality is obtained using L'Hopital's Rule. Therefore L = 2 . E
(
ln n2 + 2 n
) = lim 2n + ln 2 ¥2
n
9. First notice that if an = 0 for all n, then sn = 0 for all n and hence
¥
ås
n
converges. If
n=1
¥
¥
¥
å sn converges, then lim sn = 0 and hence
å an converges to 0. Finally, if
åa
n=1
n=1
n=1
n®¥
Page 1 of 5
n
converges
Calculus Sequences and Series Solutions
FAMAT State Convention 2015
to a nonzero number L, then lim sn = L ¥ 0 and hence
n®¥
¥
ås
n
diverges. Therefore only II is
n=1
true. B
¥
¥
1
1
10. If p £ 2 then å p
converges by limit comparison to the p-series å 2 . If p > 2
2
n=1 n + n
n=1 n
¥
¥
1
1
then å p
converges by limit comparison to the p-series å p since p > 1.
2
n=1 n + n
n=1 n
¥
1
Therefore å p
converges for all real values of p. D
2
n=1 n + n
11. Notice that an+1 = 2an . Suppose lim an = lim an+1 = L . Then L = 2L , so L2 = 2L .
n®¥
n®¥
Since a1 = 1 the limit will be nonzero and hence L = 2 .
( )
Alternatively, if a1 = 1 then a2 = 21/2 , a3 = 21/2 × 21/2
¥
lim an = 2
n®¥
¥ 2k
k=1
1
1/2
= 21/2+1/4 , etc. Then
1/2
= 21-1/2 = 2 . C
12. Using the Ratio Test, the power series will converge
2 x  3n1  n  1  lim n  1 2 x  3  2 x  3  1 , and hence x - 3 < 1 . The
when lim
n
n 
n 
2 2
n  2 2 x  3
n2
power series therefore converges on the open interval (1, 2 ) . The above limit is equal to 1
at the end points of this interval, and hence the Ratio Test is inconclusive. When x = 1 the
n
¥
-1)
(
power series is å
, which is a convergent alternating series. When x = 2 the power
n +1
n=0
¥
1
1
series is å
which diverges by limit comparison to the p-series with p = £ 1. The
2
n +1
n=0
power series therefore converges on [1, 2 ) . C
13. Let the smallest of the three angles be q . Then the other angles are q + d and q + 2d
for some positive number d. Since the sum of the angles of a triangle is 180 degrees,
3q + 3d =180° and hence q + d = 60° . Let the longest side of the triangle have length x.
Then since q + d is the second largest angle of the triangle, it is the angle between the
shortest and longest sides of the triangle. The area of the triangle is then given by
1
3
12
= 4 3. D
A = ×1× x×sin 60° = x
= 3 and hence x =
2
4
3
14. Let f ( x) = x5 . We can evaluate 10.15 be finding the Taylor series of f ( x) centered at
10. f (10) = 100000 , f ¢ (10) = 50000 , f ¢¢ (10) = 20000 , f (3) (10) = 6000 , f ( 4) (10) = 1200 ,
f (5) (10) = 120 , and f ( n) (10) = 0 for n ³ 6 . The Taylor series for f ( x) centered at 10 is
20000
6000
1200
120
f ( x) = 100000 + 50000 ( x -10) +
( x -10)2 +
( x -10)3 +
( x -10)4 + ( x -10)5
2!
3!
4!
5!
and 10.15 = f (10.1) = 1000000 + 5000 +100 +1+ 0.005 + 0.00001 = 105101.00501. D
Page 2 of 5
Calculus Sequences and Series Solutions
FAMAT State Convention 2015
1 1 1 1 1 1
¥ 1 1
¥¥ 1 1
¥
+ + + + + + = ¥1+ + + ¥¥1+ + + ¥. This is a
¥ 2 4
¥¥ 3 9
¥
2 3 4 6 8 9
¥
1
1
1
3
×
= 2 × = 3. B
product of two geometric series. å =
1
1
2
n=1 an
1- 12
3
¥
1 1 1 1 1 1 1
1 1 1 1 1
1 ¥ 1
1
16. Notice that + + + + + + + > + + + + + = å . Since å
2 3 4 6 8 9 10
2 4 6 8 10
2 n=1 n
n=1 n
¥
1
is the divergent Harmonic Series, å diverges by the Direct Comparison Test. D
n=1 bn
1
¥ 1 ¥
17. Since lim ¥= lim n = 0 , lim an = 0 by the Squeeze Theorem. A
¥
n®¥
n®¥
¥ n ¥ n®¥ 2
15. Notice that 1+
¥
18. Since
¥
å an converges, lim an = 0 and hence lim an2 = 0 . I is true. Since
åa
n=1
n=1
n®¥
n®¥
¥
alternating series,
n
¥
å an2 is also an alternating series. Since lim an2 = 0 ,
åa
n=1
n=1
n®¥
is an
n2
converges by
the Alternating Series Test. II is also true. C
t
¥
t
4
4
¥4
¥
19. Notice first of all that indeed ¥
dx = lim ¥
dx = lim ¥ arctan x¥ =
2
t®¥ p 1+ x2
t®¥ ¥p
¥1
1 p 1+ x
1
(
)
(
)
4 ¥p p ¥
4
is
- ¥= 1. Since the real-valued function f ( x) =
¥
t®¥
p
p ¥2 4 ¥
p ( x2 +1)
¥
4
positive, continuous, and decreasing on the interval [1, ¥) the series å
converges
2
n=1 p (1+ n )
by the Integral Test. A
x2 x 4
1 1
2
- we get
20. Using the identity cos x = + cos ( 2x) and the fact that cos x = 1- +
2 24
2 2
2
4
¥
1 1 ¥ ( 2x) ( 2x)
x4
cos2 x = + ¥1+
- ¥= 1- x2 + - . B
2 2¥
2
24
3
¥
4
lim [ arctant - arctan1] =
¥
¥
4
1 ¥ ¥ 1 1 1 1 1
¥1
¥ ¥ 1¥
=
2
= 2 ¥1- + - + - + ¥ = 2 ¥1+ ¥ = 3. Notice that the
¥
¥
¥
¥
2
¥
¥ ¥ 2¥
n + 2¥ ¥ 3 2 4 3 5
n=1 n + 2n
n=1 n
1
telescoping series converges because lim
= 0. C
n®¥ n + 2
21.
¥
22. I and II are true for any convergent series. Since
åb
n
n=1
converges lim bn = 0 and hence
n®¥
bn < 1 for sufficiently large n. Therefore for large n, 0 < anbn < an and hence
¥
åa b
n n
n=1
converges by the Direct Comparison Test. So III is true. Notice that the condition that
¥
¥
å an and
åb
n=1
n=1
n
are series of positive terms is needed to use the Direct Comparison Test.
Page 3 of 5
Calculus Sequences and Series Solutions
FAMAT State Convention 2015
III is not true in general (consider an = bn
( -1) , so that
=
n
¥
åa
n
¥
and
n=1
åb
n
converge by the
n=1
¥
Alternating Series Test but
åa b
n n
is the divergent Harmonic Series). D
n=1
¥
23. Since
¥
¥
å an and
å bn are series of positive terms, an < an + bn and hence
å(a
n=1
n=1
n=1
diverges by the Direct Comparison Test. If an = bn for all n, then
¥
å(a
n
n
+ bn )
- bn ) converges to 0.
n=1
1
If an = bn = then
n
¥
¥
1
anbn = å 2 is a convergent p-series. Only I is true. A
å
n=1
n=1 n
¥
{sn } be the sequence of partial sums of å an and let {tn } be the sequence of
24. Let
n=1
¥
partial sums of
å b . Then lim s
n
n®¥
n=1
¥
¥
å an and
åb
n=1
n=1
n
n
is a finite number but lim tn does not converge since
n®¥
converge and diverge, respectively. Therefore lim ( sn + tn ) and lim ( sn - tn )
n®¥
n®¥
both diverge and hence I and II both diverge. III does not need to diverge; consider an =
1
. B
n
an+1
n2 +1 1
= 2
= < 1 . Therefore
25.
an
2n +1 2
1
n2
and bn =
¥
åa
n
will converge absolutely by the Ratio Test. A
n=1
¥
xn
for all real values of x,
n=0 n!
26. Using the fact that ex = å
2 -2 x
3
xe
a65 =
f
=x
( 65)
¥
2
å
n=0
( 0) .
65!
( -2x )
3 n
n!
¥
=å
n=0
( -2)n x3n+2 .
n!
The coefficient of the x65 term of f ( x) is
Therefore f ( 65) ( 0) = an ×65!. The coefficient of the x65 term (which occurs
( -2)21 .
2 21 × 65!
when n = 21) is
Therefore f ( 0) = . C
21!
21!
27. Since the power series is centered at 3 and converges when x = 5 , the radius of
convergence is at least 2. The power series then must converge for values of x in the open
interval (1, 5). Notice that power series I-IV occur when x is equal to 1, 2, 4, and 7,
respectively. Therefore power series II and III are guaranteed to converge. Notice that
more information is needed to determine whether or not power series I converges. B
( 65)
¥
28.
åa
n
will converge if and only if lim sn exists, in which case the series will converge to
n®¥
n=1
13n2 + n -1 13
= = 3.25 . B
n®¥ 4n2 + 3n + 2
4
this limit. S= lim sn = lim
n®¥
Page 4 of 5
Calculus Sequences and Series Solutions
FAMAT State Convention 2015
¥
29. The power series
åa x
n
n
is centered at 0. There are three possible types of intervals of
n=0
convergence: 1. The power series converges only when x = 0 ; 2. The power series
converges for all real values of x; or 3. The power series converges on an interval of finite
length centered at 0 (and possibly including one, both, or none of the endpoints). None of
the choices fall into any of these three categories. E
30. A
Page 5 of 5