CAS MA 124 Study Guide - Midterm Guide: Divergent Series, Alternating Series, Convergent Series
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10 Oct 2018
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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 k≥1 for klarge.
ln k
k≥1
kfor klarge.
∞
X
k=1
ln k
k≥
∞
X
k=1
1
k
∞
X
k=1
1
kdiverges since p−series 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
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)·x−ln x
x2=1−ln 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.
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
ET≤k·(b−a)
12 ·(∆x)2ES≤K·(b−a)
180 ·(∆x)4
where |f(2)(x)| ≤ kon [a, b] and |f(4)(x)| ≤ Kon [a, b] and ∆x= (b−a)/n in both cases.
If T(8) and S(8) are used to approximate
Z1
0
e−xdx
give upper bounds for ETand ESusing the above formulas.
ANSWER: We first note that
∆x=1
8
We next calculate derivatives:
f0(x) = −e−x⇒f(2)(x) = e−x⇒f(3)(x) = −e−x⇒f(4)(x) = e−x
So both |f(2)(x)|and |f(4)(x)|=|e−x|=e−x. Since
e−x=1
ex
and exis increasing, we have e−xis largest on the interval [0,1] when exis smallest which
occurs for x= 0. So
|f(2)(x)|=|f(4)(x)|=1
ex≤1
e0= 1
Hence
k=K= 1
Inserting these values and the value for ∆xinto the expressions for ETand ESwe get
ET≤1·1
12 ·1
82=1
12 ·82
ES≤1·1
180 ·1
84=1
180 ·84