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Solved Exam 2 - The Cauchy Condensation | MUIN 425, Exams of Music

Material Type: Exam; Class: Live Music Production and Promotion; Subject: Music Industry; University: University of Southern California; Term: Fall 2009;

Typology: Exams

2009/2010

Uploaded on 02/24/2010

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The Cauchy Condensation Test
Dave L. Renfro
If 0an+1 anfor each n, then
1
X
n=1
an<1 ()
1
X
n=1
2na2n<1.
This is called the Cauchy Condensation Test, and the associated series
1
X
n=1
2na2nis
often called the condensed series.1Note that we are not claiming the condensed series has the
same sum as the original series.
Proof: The proof is based on the “grouping method” for proving the divergence of the harmonic
series. Assume that 0an+1 anfor each n. We must prove that the biconditional
1
X
n=1
an<1 ,
1
X
n=1
2na2n<1holds. This will be done by proving the two associated
conditional statements ()and () separately.
())Assume
1
X
n=1
an<1. Then
a1+a2+ (a3+a4)+(a5+a6+a7+a8)+(a9+a10 ++a16) +  <1
=)a1+a2+ (a4+a4) + (a8+a8+a8+a8) + (a16 +a16 ++a16) + <1
=)a1+a2+ 2a4+ 4a8+ 8a16 + <1:
Doubling each term of this convergent series gives the following convergent series:
2a1+ 2a2+ 4a4+ 8a8+ 16a16 + <1:
Since the series above equals 2a1+
1
X
n=1
2na2n, it follows that
1
X
n=1
2na2n<1.
(()Assume
1
X
n=1
an=1. Then
a1+ (a2+a3)+(a4+a5+a6+a7)+(a8+a9++a15) +  =1
=)a1+ (a2+a2)+(a4+a4+a4+a4)+(a8+a8++a8) +  =1
=)a1+ 2a2+ 4a4+ 8a8+ =1:
Since the series above equals a1+
1
X
n=1
2na2n, it follows that
1
X
n=1
2na2n=1.2
1More g enerall y, the followin g holds. If 0an+1 anfo r each nand fnkgis a s ubsequ ence of th e posit ive
intege rs such tha t M > 0ex ists for wh ich nk+1 nkM(nknk1)(i.e. the gap s betwee n nkand nk+1 incr ease
at mos t geome trically ), then
1
X
n=1
an<1 ()
1
X
k=1
(nk+1 nk)ank<1:
Putt ing fnkg1
k=1 =2k1
k=1 gives th e Cauchy C ondens ation Test.
2Wha t we have done is t o prove P1
n=1 an=1 ) P1
n=1 2na2n=1, which is t he contrapos itive (u nder
the ass umpti on that 0an+1 an) of P1
n=1 2na2n<1 ) P1
n=1 an<1.
pf2

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The Cauchy Condensation Test

Dave L. Renfro

If 0  an+1  an for each n, then

X^1

n=

an < 1 ()

X^1

n=

2 na 2 n < 1.

This is called the Cauchy Condensation Test, and the associated series

X^1

n=

2 na 2 n is

often called the condensed series.^1 Note that we are not claiming the condensed series has the same sum as the original series.

Proof: The proof is based on the ìgrouping methodî for proving the divergence of the harmonic

series. Assume that 0  an+1  an for each n. We must prove that the biconditional X^1 n=

an < 1 ,

X^1

n=

2 na 2 n < 1 holds. This will be done by proving the two associated

conditional statements () and () separately.

()) Assume

X^1

n=

an < 1. Then

a 1 + a 2 + (a 3 + a 4 ) + (a 5 + a 6 + a 7 + a 8 ) + (a 9 + a 10 +    + a 16 ) +    < 1 =) a 1 + a 2 + (a 4 + a 4 ) + (a 8 + a 8 + a 8 + a 8 ) + (a 16 + a 16 +    + a 16 ) +    < 1 =) a 1 + a 2 + 2a 4 + 4a 8 + 8a 16 +    < 1 : Doubling each term of this convergent series gives the following convergent series:

2 a 1 + 2a 2 + 4a 4 + 8a 8 + 16a 16 +    < 1 :

Since the series above equals 2 a 1 +

X^1

n=

2 na 2 n , it follows that

X^1

n=

2 na 2 n < 1.

(() Assume

X^1

n=

an = 1. Then

a 1 + (a 2 + a 3 ) + (a 4 + a 5 + a 6 + a 7 ) + (a 8 + a 9 +    + a 15 ) +    = 1 =) a 1 + (a 2 + a 2 ) + (a 4 + a 4 + a 4 + a 4 ) + (a 8 + a 8 +    + a 8 ) +    = 1 =) a 1 + 2a 2 + 4a 4 + 8a 8 +    = 1 :

Since the series above equals a 1 +

X^1

n=

2 na 2 n , it follows that

X^1

n=

2 na 2 n = 1.^2

(^1) More generally, the following holds. If 0  an+1  an for each n and fnk g is a subsequence of the positive integers such that M > 0 exists for which nk+1 nk  M (nk nk 1 ) (i.e. the gaps between nk and nk+1 increase at most geometrically), then X^1 n=

an < 1 () X^1 k=

(nk+1 nk ) ank < 1 :

Putting fnk g^1 k=1 =  2 k^1 k=1 gives the Cauchy Condensation Test. (^2) What we have done is to prove P (^1) n=1 an = 1 ) P (^1) n=1 2 na 2 n = 1 , which is the contrapositive (under the assumption that 0  an+1  an) of P^1 n=1 2 na 2 n^ < 1 ) P^1 n=1 an < 1.

EXAMPLES^3

X^1

n=

n^2 The condensed series is

X^1 n=

2 n^ 

(2n)^2

X^1

n=

2 n

X^1

n=

n ;

which is a convergent geometric series.

X^1

n=

n The condensed series is X^1 n=

2 n^ 

(2n)

X^1

n=

which is clearly divergent.

X^1

n=

np^ The condensed series is

X^1 n=

2 n^ 

(2n)p

X^1

n=

2 nnp

X^1

n=

21 p

n ;

which is a geometric series with r = 2^1 p. This diverges when jrj  1 , 21 p^  1 , 1 p  0 , p  1 ; and converges when jrj < 1 , 21 p^ < 1 , 1 p < 0 , p > 1 :

X^1

n=

n(ln n)p^ The condensed series is

X^1

n=

2 n^ 

2 n^ (ln 2n)p

X^1

n=

(n ln 2)p

X^1

n=

np^ (ln 2)p

ln 2

p (^) X 1

n=

np^

which (use the previous example) converges when p > 1 and diverges when p  1.

X^1

n=

n ln n[ln(ln n)]p^ The condensed series is

X^1

n=

2 n^ 

2 n^  (ln 2n)  [ln (ln 2n)]p

X^1

n=

(n ln 2) [ln (n ln 2)]p

X^1

n=

(ln 2)  n [ln n + ln (ln 2)]p

which converges when p > 1 (direct comparison with the previous example) and diverges when p  1 (limit comparison with the previous example). (^3) In the examples that follow, observe how a use of the Cauchy Condensation Test transforms the series into one that converges (or diverges) more rapidly than the original series. In particular, the Cauchy Condensation Test transforms a p series into a geometric series (hence, these two types can be considered as relatives of each other) and it transforms an níth order logarithmic p series into an (n 1)íst order logarithmic p series.