NYCU 0413 Calculus Final Exam 2022
Problem 1. Use the test you learned to determine whether the series is convergent or divergent.
(1)
(2)
(3)
8
ÿ
1
nln n
n=1
;
8
ÿ
1
(HINT: the Integral Test);
n ln n(ln ln n)p
n=2
8
ÿ
π
(1 ´ cos ).
n
n=1
?
(
2 + (´1)n )n
;
(4)
n3
3n
n=1
8
ÿ
(5)
8
ÿ
1
;
(ln n)ln n
n=2
8
ÿ
(6)
1
?
;
n ln n
n=2
(7)
8
ÿ
1
, where a ą 1;
ln n
a
n=1
(8)
8
ÿ
ln n
;
n
2
n=1
8
ÿ
1
? , where a ą 1;
(9)
a n
n=1
(10)
8
ÿ
n tan
n=1
π
2n+1
;
8
ÿ
1
1
(11)
sin ;
p
n
n
n=1
(12) For x ě 0,
(13)
(14)
8
ÿ
xn
;
(1 + x)(1 + x2 ) ¨ ¨ ¨ (1 + xn )
n=1
8
ÿ
3n n!
;
n
n
n=1
8
ÿ
?
?
4
( n + 1 ´ n2 + n + 1);
n=1
8 ż 1/n ?
ÿ
x
(15)
dx;
1 + x2
n=1 0
(16)
8
ÿ
en n!
;
n
n
n=1
8
ÿ
2n n!
(17)
;
nn
n=1
8
ÿ
2n
(18)
;
3ln n
n=1
8
ÿ
1 ¨ 3 ¨ 5 ¨ ¨ ¨ (2n ´ 1)
(19)
;
n!
n=1
(20)
8
ÿ
sin
n=1
(21)
8
ÿ
(
n=1
(22)
(23)
1
;
n2
n
)n ;
2n + 1
8
ÿ
ln n
;
2
n
n=2
8
ÿ
1
n=1
nln n
;
8
ÿ
ln n n
(
) ;
(24)
n
n=2
8
ÿ
(α + 1)(2α + 1) ¨ ¨ ¨ (nα + 1)
;
(25)
(β + 1)(2β + 1) ¨ ¨ ¨ (nβ + 1)
n=1
8
ÿ
(26)
1
?
;
3
n2 ´ 1
n=2
(27)
8
ÿ
1
1
ln(1 + );
n
n
n=2
?
8
ÿ
(´1)n+1 n
?
;
(28)
3
n
n=2
(29)
(30)
8
ÿ
n+1
;
ln(n2 + 1)
n=2
8
ÿ
(´1)n
n=1
(31)
8
ÿ
cos nπ
;
n
1
a
?
? ;
n(n + 1)( n + 1 + n)
n=1
8
ÿ
(32) For p ą 0,
(e ´ (1 +
n=1
(33)
1 n p
) );
n
8
ÿ
1
;
n ln n ¨ (ln ln n)p
n=2
8
ÿ
1 3 + x 2n
(34)
(
) ;
2n 3 ´ 2x
n=1
(35)
8
ÿ
xn
?
;
n+1
n=1
8
ÿ
(36)
ln n
;
2n2 ´ 1
n=2
(37)
8
ÿ
cos nπ
;
n
+
1
n=0
(38)
8
ÿ
sint
n=1
2n ´ 1
πu.
2
Problem 2. For α ą 0. Suppose that the series
8
ÿ
a2n
converges. Show that
n=1
absolutely convergent.
Problem 3. Use the test you learned to determine whether the series
(1)
8
ÿ
(´1)n
n=1
n´1
? ;
(n + 1) ¨ 100 n
8
ÿ
(2) For a ą 0,
(´1)n´1
n=1
(3)
8
ÿ
(´1)n´1
n=1
(4)
(5)
1
1
np+ n
;
8
ÿ
ln(2 + n1 )
(´1)n´1 a
;
(3n
´
2)(3n
+
2)
n=1
8
ÿ
(´1)n
n=1
1 + 12 + 13 + ¨ ¨ ¨ + n1
;
n
8
ÿ
(x ´ 2)2n
(6)
;
4n n
n=1
(7)
1
1
¨
;
n 1 + an
8
ÿ
1
(´1)n+1 ? ;
n
n=1
8
ÿ
a
? n
is
2
n + α2
n=1
(8)
8
ÿ
(´1)n´1
n=1
2n
;
n2
8
ÿ
1
2n ´ 1
(9)
sint
πu;
n
2
n=2
(10)
8
ÿ
?
?
(´1)n ( n + 1 ´ n);
n=1
8
ÿ
cos nπ
(11)
;
n2
n=1
8
ÿ
(12)
(´1)n
;
(2n + 1)!
n=0
(13)
8
ÿ
(´1)n ln(n + 1)
;
n
+
1
n=1
(14)
8
ÿ
(´1)n+1 tan´1 n.
n=1
is absolutely convergent, conditionally convergent, or divergent.
Problem 4. Find the interval of convergent of the following power series
8
ÿ
3n + (´2)n
(1)
(x + 1)n ;
n
n=1
(2)
(3)
(4)
8
ÿ
ln(n + 1) n+1
x ;
n
+
1
n=1
8
ÿ
1
1 ´ x 2n
(
) ;
3n
+
1
1
+
x
n=0
8
ÿ
n22n xn (1 ´ x)n ;
n=1
(5)
8
ÿ
1
?
xn ;
n
+
1
n=0
8
ÿ
(x ´ 3)n
;
(6)
3n n
n=1
(7)
8
ÿ
(´1)n
n=1
(8)
8
ÿ
(sin
n=1
(x ´ 4)n
;
n
3+x n
1
)¨(
) ;
3n
3 ´ 2x
(9)
(10)
(11)
8
ÿ
ln(n + 1) n+1
x ;
n
+
1
n=1
8
ÿ
1
xn
¨
;
n + (´2)n
3
n
n=1
8
ÿ
n+1 (x
(´1)
n=0
(12)
8
ÿ
´ 1)n+1
;
n+1
n´1 (2x
(´1)
n=1
´ 4)n
.
n
Problem 5. Representations of the following functions
(1) f (x) = ln(1 + x + x2 + x3 ); (HINT: 1 ´ x4 )
(2) f (x) = sin2 x;
(3) f (x) = (1 + x) ln(1 + x);
(4) f (x) =
12 ´ 5x
;
6 ´ 5x ´ x2
(5) f (x) =
3x
;
2 ´ x ´ x2
(6) f (x) =
x2
3 ´ 2x
with center ´4;
+ 3x + 2
(7) f (x) = tan´1 (
1 ´ 2x
);
1 + 2x
(8) f (x) = cos2 (2x);
(9) f (x) =
1+x
;
(1 ´ x)2
x
(10) f (x) = ?
;
π(1 ´ x)2
(11) f (x) =
´2x2
;
(1 ´ x)2
(12) f (x) =
πx2
.
(1 ´ x)2
as power series.
Problem 6. True or False.
8
8
ÿ
ÿ
?
(1) If
an is also divergent.
an is divergent, then
n=1
(2) Let an ą 0 for n P N. If
n=1
8
ÿ
8
ÿ
?
an is also convergent.
an is convergent, then
n=1
n=1
8
ÿ
(3) There exists a power series
an (x ´ k)n whose interval of convergence is [0, 8).
n=0
8
ÿ
an+1
(4) If an ą 0, n ě 1, lim an = L and L P Rzt0u, then
is divergenet.
nÑ8
an
n=0
8
ÿ
(5) The domain of
an (x ´ 1)n cannot be exactly the interval (0, 3).
n=0
8
ÿ
(6) If the series
a2n
n=1
8
ÿ
an
is convergent (an ą 0), then the series
is also convergent.
n
n=1
8
ÿ
(7) If an ě 0 for n P N and lim nan = 1, then
nÑ8
8
ÿ
(8) If an ě 0 for n P N and lim nan = 0, then
nÑ8
nÑ8
8
ÿ
8
ÿ
(10) If an ě 0 for n P N and lim n an = 0, then
nÑ8
(11) If
an and
n=1
(12) If f (x) =
an is convergent.
n=1
bn both converge, then
n=1
8
ÿ
an is convergent.
n=1
2
8
ÿ
an is convergent.
n=1
(9) If an ě 0 for n P N and lim n2 an = 1, then
8
ÿ
an is convergent.
n=1
8
ÿ
an bn converges.
n=1
n
an x converges for |x| ă 2, then
ż1
f (x) dx =
0
n=0
8
ÿ
(13) If the power series
8
ÿ
an
.
n+1
n=0
an xn converges for x = 2, then it also converges for x = ´2.
n=1
(14) If f (x) =
8
ÿ
(´1)n [
n=0
(15) If
8
ÿ
an is convergent, then
n=1
(16) If
8
ÿ
?
x2n
x2n+1
π
+
], then f ( ) = 2.
(2n)! (2n + 1)!
4
8
ÿ
a2n is also convergent.
n=1
8
ÿ
?
an is also convergent.
an (x ´ 1) is convergent for x = 0, then
n
n=0
(17) If
8
ÿ
n=0
u2n
and
n=1
(18) If
8
ÿ
n=1
8
ÿ
vn2
are convergent, then
n=1
u2n
and
8
ÿ
n=1
8
ÿ
(un + vn )2 is also convergent.
n=1
vn2
are convergent, then
8
ÿ
n=1
|un vn | is also convergent.
(19) If
8
ÿ
|un vn | is convergent, then
n=1
(20) If
8
ÿ
8
ÿ
(un + vn )2 is also convergent.
n=1
un is convergent and un ě vn , then
n=1
8
ÿ
vn is also convergent.
n=1
?
3
3n2 + 5
Problem 7. What value of k can make the series
converge?
nk
n=1
8
ÿ
Problem 8. For what values of the positive constants a and b does the following series converge
absolutely? For what values does it converge conditionally?
a´
b a b a b
+ ´ + ´ + ¨¨¨
2 3 4 5 6
Problem 9. Suppose that a1 = 2 and for all n P N,
1
1
an+1 = (an + ) .
2
an
Prove the following statements.
(1) The limit lim an exists;
nÑ8
(2) The series
8
ÿ
(
n=1
an
´ 1) is convergent.
an+1
Problem 10. Suppose that an ą 0 for all n P N and
lim n(
nÑ8
Prove the series
8
ÿ
an
´ 1) = a ą 0.
an+1
(´1)n´1 an is convergent.
n=1
Problem 11. Suppose that the interval of convergence of
an (x ´ 3)n is (´5, 11). Which the
n=1
following statement is True?
(A) The radius of convergence of
8
ÿ
8
ÿ
nan (x + 2)3n is 8;
n=1
(B) The series
8
ÿ
nan (x + 2)3n is convergent if x P (´4, 0);
n=1
(C) The series
8
ÿ
n(n ´ 1)an (x + 2)3n is convergent if x P [´4, 0];
n=1
(D) The series
8
ÿ
n=1
an (x + 2)3n is convergent if x P (´4, 0).
Problem 12. If the series
an (x + 3)n is convergent on [´5, 1], then the series
n=1
convergent on
(A) (´5, 1);
8
ÿ
(B) (´3, 5);
Problem 13. Let S(x) =
statement is True?
(D) [´1, 5].
8
8
ÿ
ÿ
n2 n
n
x . Suppose that g(t) =
tn´1 . Which the following
n!
(n
´
1)!
n=1
n=1
n
(A) If g(t) =
tn´1 , then
(n
´
1)!
n=1
żx
g(t) dt = x
0
8
ÿ
xn
= xex .
(n
´
1)!
n=1
d
(xex ).
dx
(C) S(x) = x(x + 1)ex if x P (´8, 8).
(D) S(x) = x
nan (x ´ 1)n is
n=1
(C) (´1, 3);
8
ÿ
(B) g(x) =
8
ÿ
8
ÿ
n
xn´1 .
(n ´ 1)!
n=1
Problem 14. Given the function f (x) = ex . Which the following statement is True?
8
ÿ
f (n+1) (0) n+1
(A) The Maclaurin series of f is
x ;
(n + 1)!
n=´1
8
ÿ
f (n) (0) n
(B) The Maclaurin series of f is
x ;
n!
n=0
(C) The radius of convergence of Maclaurin series of f is 8;
(D) When x = 0, then Maclaurin series of f converges to 0.
Problem 15. Discuss the convergence/divergence of the series
8
ÿ
(1 ´
n=1
ln n n
) .
n