CAS MA 124 Study Guide - Midterm Guide: Divergent Series, Alternating Series, Convergent Series

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10 Oct 2018
Professor
MA 124 - Fall ’12 - Exam 2A - ANSWERS
1) a) (4 points) True or False:
If limk→∞ ak= 0 then P
k=1 akconverges.
b) (6 points) Does the following series converge or diverge ? Explain your answer.
X
k=1
ln k
k
c) (6 points) Does the following series converge or diverge ? Explain your answer.
X
k=1
(1)kln k
k
ANSWER:
a) FALSE
b) Diverges.
Using Comparison Test we note:
ln k1 for klarge.
ln k
k1
kfor klarge.
X
k=1
ln k
k
X
k=1
1
k
X
k=1
1
kdiverges since pseries with p= 1
Our series bigger than a divergent series, hence diverges by Comparison Test.
NOTE: One can also use the Integral Test to show this series diverges by showing the
improper integral
Z
1
ln x
xdx
diverges.
Answer to c) on next page
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c) Converges
Using the Alternating Series Test we note:
1)
lim
k→∞
ln k
k= 0
since
lim
x→∞
ln x
xis indeterminate form
Applying L’Hopital’s Rule
lim
x→∞
ln x
x= lim
x→∞
(ln x)0
x0= lim
x→∞
1
x= 0
2)
ak+1 =ln(k+ 1)
(k+ 1) ln k
k=ak
since if we let
f(x) = ln x
x
we have
f0(x) = (ln x)0·x(ln x)·x0
x2=(1/x)·xln x
x2=1ln x
x2
Since 1 ln x < 0 for xlarge we have f0(x)<0 for xlarge, hence f(x) is decreasing,
which implies
ak+1 =f(k+ 1) f(k) = akfor klarge
Because 1) and 2) are satisfied the series converges by the Alternating Series Test.
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2) (16 points) If the Trapezoid rule T(n) and Simpson’s Rule S(n), with n-subintervals,
are used to find an approximate value of
Zb
a
f(x)dx
then the errors ETand ESsatisfy the inequalities
ETk·(ba)
12 ·(∆x)2ESK·(ba)
180 ·(∆x)4
where |f(2)(x)| ≤ kon [a, b] and |f(4)(x)| ≤ Kon [a, b] and ∆x= (ba)/n in both cases.
If T(8) and S(8) are used to approximate
Z1
0
exdx
give upper bounds for ETand ESusing the above formulas.
ANSWER: We first note that
x=1
8
We next calculate derivatives:
f0(x) = exf(2)(x) = exf(3)(x) = exf(4)(x) = ex
So both |f(2)(x)|and |f(4)(x)|=|ex|=ex. Since
ex=1
ex
and exis increasing, we have exis largest on the interval [0,1] when exis smallest which
occurs for x= 0. So
|f(2)(x)|=|f(4)(x)|=1
ex1
e0= 1
Hence
k=K= 1
Inserting these values and the value for ∆xinto the expressions for ETand ESwe get
ET1·1
12 ·1
82=1
12 ·82
ES1·1
180 ·1
84=1
180 ·84
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