CAUCHY CONDENSATION PRINCIPLE
JIA-MING (FRANK) LIOU
Let (an ) be a sequence of real numbers such that
0 ≤ an+1 ≤ an
for all n ≥ 1, i.e. (an ) is a nonnegative, non increasing sequence of real numbers.
Lemma 0.1. Suppose that 2k ≤ n < 2k+1 . Then
a1 +
k
n
k
j=1
i=1
j=1
X
X
1X j
2 a2j ≤
ai ≤ a1 +
2j a2j .
2
Proof. Since (an ) is nonincreasing, we observe that a3 + a4 ≥ 2a4 , a5 + a6 + a7 + a8 ≥ 4a8
and
a2k−1 +1 + a2k−1 +2 + · · · + a2k ≥ 2k−1 a2k .
Since 2k ≤ n < 2k+1 , and ai ≥ 0 for all i, we know
n
X
ai = a1 + a2 + a3 + a4 + · · · + a2k−1 +1 + · · · + a2k + a2k +1 + · · · + an
i=1
≥ a1 + a2 + a3 + a4 + · · · + a2k−1 +1 + · · · + a2k
= a1 + a2 + (a3 + a4 ) + · · · + (a2k−1 +1 + · · · + a2k )
≥ a1 + a2 + 2a4 + · · · + 2k−1 a2k−1
1
= a1 + (2a2 + 4a4 + · · · + 2k a2k )
2
k
1X j
= a1 +
2 a2j .
2
j=1
Similarly, we observe that by the fact that (an ) is nonincreasing, a2 + a3 ≤ 2a2 , a4 + a5 +
a6 + a7 ≤ 4a4 and
a2k + a2k +1 + · · · + a2k+1 −1 < 2k a2k .
Since ai ≥ 0, we know that
n
X
ai = a1 + · · · + a2k−1 + · · · + an
i=1
≤ a1 + a2 + a3 + · · · + a2k + a2k +1 + · · · + an + an+1 + · · · + a2k+1 −1
= a1 + (a2 + a3 ) + · · · + (a2k + a2k +1 + · · · + a2k+1 −1 )
≤ a1 + 2a2 + · · · + 2k a2k
= a1 +
k
X
2j a2j .
j=1
Hence we proved the inequality.
1
2
JIA-MING (FRANK) LIOU
Corollary 0.1. (Cauchy condensation theorem)
be a nonnegative non increasing
P Let (an ) P
∞
n
sequence of real numbers. Then either both ∞
a
and
n=1 n
n=1 2 a2n converge or diverge.
Proof. The proof follows from the above inequality. ( The proof is similar to the comparison
test.)
Theorem 0.1. The series
∞
X
1
is convergent if p > 1 and divergent if p ≤ 1.
np
n=1
Proof. Let an =
1/np .
Then
n
2 a2n
Hence
1
1
= 2 · n p = np−n =
(2 )
2
n
∞
X
n=1
2n a2n =
1
2p−1
n
.
n
∞ X
1
2p−1
n=1
is a geometric series with ratio 1/2p−1 . If p > 1, P
p − 1 > 0. In this case 2p−1 > 1 and
p−1 )n is convergent. If p < 1,
p−1
thus 1/2
< 1. We know that theP
infinite series ∞
n=1 (1/2
∞
1/2p−1P> 1, then the infinite series n=1 (1/2p−1 )n is divergent. If p = 1, 1/2p−1 = 1. We
n
know ∞
n=1 1 is divergent.
Similarly, you can determine when the following series is convergent using the Cauchy
condensation theorem
∞
X
1
,
n(ln n)p
n=2
1 n
where ln n = loge n, and e = lim 1 +
.
n→∞
n