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(MATH024)004final_solution_2006.pdf
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Math 004 -Advanced Mathematics II
Final Examination, Spring Semester, 2006

Time Allowed: 3 Hours Total Marks: 100
Note: Calculators are not allowed. Please write details to receive marks. Only the following formulas can be used:
n+1
xx
e x dx = e + C;
x n dx =+C, (n =..1);
n +1
dx

dx = arctan x + C;

= ln |x| + C;

x2 +1
dx
= arcsin x + C;
sin x dx = . cos x + C;

cos x dx = sin x + C;

1 . x2 (.1)dx

= arccos x + C;

1 . x2
.
.
2n.1
(.1)n.1x
1

n
, |x| < 1; sin x =
|x| < +;

= x

n=0
,

1 . x

(2n . 1)!

n=1
.
.
2n
(.1)nx
n
x
x
|x| < +;

cos x =

|x| < +.

=

e

,

,

n!

(2n)!

n=0
n=0
1. (20 Marks) Determine whether the following series are convergent

. 1/n
.
sin n

.
.t2
(a) ; (c)

dt;

e

n2
0 n=1 n=1
. 1/n
..
1

100 .t2
e

(b) ; (d)

dt.

t

n2 + 1 ln(n2 + 1)
0 n=1 n=1
Solution Since

sin n

1
,
2 2
n

n

. 1
and nis convergent, by the Comparison Test, the series (a) is con-
2
vergent.
Since
1

n2+1 ln(n2+1) 1
lim = ,
n
n ln n 2
so the convergence of the series (b) is the same as the series 1 .
n ln n
The later series is divergent, by the Integral Test, since
. A
dx
+, as A +.
2 x ln x
Hence the series (b) is divergence.

Since . x
.t2

e dt .x
e2 1
lim 0 = lim = ,
x0 x x0 22
thus, we know that

. 1/n .t2
e dt
1
0
lim = .
x0 1/n 2
By the Comparison Test, the series (c) is divergent.

Since .
x t100 e .t2 dt 100 .x
xe2 1
lim 0 = lim = ,
101 100
x0 xx0 101 x101
thus, by the Comparison Test, the series (d) is convergent.

2. (30 Marks) Suppose f is a continuous function with 0 <m f(x) M in [0, +). Discuss the convergence of the series:
. . n
. n .
(a) f(t)dt; (d) (.1)n f(t)dt;
00
n=1 n=1

.. 1/n .. 1/n
(b) f(t)dt; (e) (.1)n f(t)dt;
00
n=1 n=1

.. 1/n2 .. 1/n2
(c) f(t)dt; (f) (.1)n f(t)dt.
00
n=1 n=1
Solution Since f(x) is continuous, with 0 <m f(x) M, we know that .
n
f(x)dx m n,
0
which implies that
n
f(x)dx +,
0
as n . Thus, the series (a) and (d) are divergent, since the n-th terms do not have zero limit as n approaches .
The integral
. 1/n
m
f(t)dx ,
0 n
by the Comparison Test, the series (b) is divergent, since m is diver-
n
gent.
On the other hand, since the same integral

. 1/n
M
0 f(t)dx ,
0 n
we have
. 1/n
lim f(t)dx =0.
n
0
It is deceasing in n:
. 1/n . 1/(n+1)
f(t)dx f(t)dx.
00
We know that the series (e) is convergent. A similar argument applies to the series (f). So (f) is also convergent.
Finally, since
. 1/n2
M
f(t)dx ,
n2
0
. M
and nis convergent, by the Comparison Test, we know that the
2
series (c) is convergent.
In summary, the series (c), (e) and (f) are convergent, while the series (a