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

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10 Oct 2018

School

Department

Course

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 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