MATH 263C, Winter 2007, Midterm 1
Name: SOLUTION
1. (5 points) Find a series §1
n=1 an with limn!1 an = 0 but the series is divergent.
1
1 p1
ANSWER: The series §1
n=1 n or §n=1 n have this property, because they are
divergent as p-series with p less than or equal to 1, and limn!1 an = 0.
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2. Use any kind of TESTs you know to show if each of the following series is divergent or convergent (show your work).
1
(a) (5 points) §1
n=2 nln(n)
ANSWER: As we know, n > ln(n) hence n2 > nln(n). This shows that
1
1
1
1
1
1
n2 < nln(n) . Moreover these two series §n=2 nln(n) and §n=2 n2 have positive terms.
1
While §1
n=2 n2 is a convergent p series with p = 2 > 1, we cannot say anything about
1
convergence or divergence of §1
n=2 nln(n) , see the Comparison Test. Thus we have to
use other test. We are thinking about the Integral Test, and to use this test we have
1
to prove ¯rst that the function f (x) = xln(x)
is positive, continuous, and decreasing in
[2; 1)). For continuity and the positivity it is clear. For the decreasing of f(x) we need
to to see at xln(x) < (x + 1)ln(x + 1) ) f (x) > f (x + 1). Thus the Integral Test can
be checked to use.
R1 1
dx = limx!1 [ln(ln(x)) ¡ ln(ln2)] = 1, doesn't converge. Thus the
1 xln(x)
given series is divergent.
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(b) (5 points) §1
n=1
sin2 [(2n¡1) ¼
2]
p
n3
ANSWER: Since 2n ¡ 1 is an odd number, sin[(2n ¡ 1) ¼2 ] is either 1 or -1.
Therefore sin2 [(2n ¡ 1) ¼2 ] = 1, and so
2
sin
§1
n=1
By p-series theorem with p =
[(2n ¡ 1) ¼2 ]
1
p
= §1
n=1 3 :
3
n2
n
3
2
we see that our series is convergent.
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2¡cos(n)
(c) (6 points) §1
n=1
n2
Typeset by AMS-TEX
1
2
3
1
1
ANSWER: We know that 3 ¸ 2 ¡ cos(n) for all n. Hence §1
n=1 n2 = 3§n=1 n2 ¸
2¡cos(n)
1
§1
. Moreover the terms of these two series are positive and 3§1
n=1
n=1 n2 is
n2
2¡cos(n)
convergent as a p-series with p = 2. Hence by the Comparison Test, §1
is
n=1
n2
convergent.
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3. Use the Integral Test to check if the following series are convergent (show your
work).
¡n
(a) (6 points) §1
n=1 ne
2
ANSWER: To be able to use the Integral Test, we have to check ¯rst if the
2
function f (n) = an = ne¡n is positive, continuous and decreasing in [1; 1). It is easy
to see that the function is positive and continuous in [1; 1). For the decreasing we look
2
2
2
at the ¯rst derivative of f (x); f 0 (x) = e¡x ¡ 2x2 e¡x = ¡(2x2 + 1)e¡x < 0 for all
x 2 [1; 1), and this shows that f (x) is decreasing in the interval [1; 1). Now we can
evaluate the integral to see our answer:
Z 1
2
2
1
1
1
1
xe¡x dx = limx!1 [¡ e¡x + e¡1 ] = 0 +
= :
2
2
2e
2e
x=1
2
¡n
Hence by the integral Test, the series §1
is convergent.
n=1 ne
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¡n
(b) (6 points) §1
n=1 e
ANSWER: We can easily prove that the function f(x) = e¡x is positive, continuous and decreasing (proof is as (a) before, leaving it for you as HW!). Now evaluate
the integral (Math 263B):
Z 1
e¡x dx = limx!1 [¡e¡x + e¡1 ] = 0 + e¡1 = e¡1 :
x=1
This integral converges, hence the series is convergent.
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4. Show if the following alternating series are convergent or divergent (show work)
n¡1 ln(n)
(a) (6 points) §1
n=1 (¡1)
n
n¡1 ln(n)
1
n¡1 ln(n)
ANSWER: §1
= 0 ¡ ln2
n=1 (¡1)
n
2 + §n=3 (¡1)
n . Hence the series
1
n¡1 ln(n)
1
§n=1 (¡1)
is converegent if and only if the series §n=3 (¡1)n¡1 ln(n)
is convern
n
gent. Therefore we consider only for n ¸ 3
3
Let bn =
ln(n)
n
then bn > 0 for all n ¸ 3. The function f (x) =
0
continuous. Now we look at the derivative: f (x) =
1
x x¡(lnx)(1)
x2
1
n
= limn!1
for x ¸ 3; ln(x) > 1. Now limn!1 ln(n)
n
have shown that the following 2 conditions hold:
(i) bn > bn+1 for all n ¸ 3, and
(ii) limn!1 bn = 0:
1
=
= 1¡lnx
x2
1
limn!1 n =
lnx
x
is also
< 0 because
0. Hence we
By the Alternating Test, the series is convergent.
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¼
n
(b) (6 points) §1
n=2 (¡1) sin( n )
ANSWER: Let bn = sin( ¼n ) then bn > 0 for all n ¸ 2. Hence the series is an
alternating series. We know from the trigonometric functions, that sin( n¼ ) is continuous
and decreasing in [2; 1). Moreover, limn!1 sin( n¼ ) = 0: Hence the series is convergent.
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n¡1
5. (5 points) Find all values of r that make the formula §1
=
n=1 ar
(Use a theorem in the book, what we often discussed in class.)
a
1¡r )
a
1¡r
correct.
ANSWER: This series is a geometric series. Hence it is convergent (and equals
when jrj < 1. This is the answer of this question.