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.
