2k+1, and ai≥ 0 for all i, we know n X i=1 ai = a1+ a2+ a3+ a4

CAUCHY CONDENSATION PRINCIPLE

JIA-MING (FRANK) LIOU

Let (a_n) be a sequence of real numbers such that 0 ≤ a_n+1≤ a_n

for all n ≥ 1, i.e. (an) is a nonnegative, non increasing sequence of real numbers.

Lemma 0.1. Suppose that 2^k ≤ n < 2^k+1. Then a1+1

2

k

X

j=1

2^ja₂^j ≤

n

X

i=1

ai ≤ a₁+

k

X

j=1

2^ja₂^j.

Proof. Since (an) is nonincreasing, we observe that a3+ a4 ≥ 2a₄, a5+ a6+ a7+ a8 ≥ 4a₈ and

a₂k−1+1+ a₂k−1+2+ · · · + a₂k ≥ 2^k−1a₂k. Since 2^k≤ n < 2^k+1, and a_i≥ 0 for all i, we know

n

X

i=1

a_i = a₁+ a₂+ a₃+ a₄+ · · · + a₂k−1+1+ · · · + a₂k+ a₂k+1+ · · · + a_n

≥ a₁+ a₂+ a₃+ a₄+ · · · + a₂k−1+1+ · · · + a₂k

= a1+ a2+ (a3+ a4) + · · · + (a₂^k−1₊₁+ · · · + a₂^k)

≥ a₁+ a2+ 2a4+ · · · + 2^k−1a₂^k−1

= a₁+ 1

2(2a₂+ 4a₄+ · · · + 2^ka₂k)

= a₁+ 1 2

k

X

j=1

2^ja₂^j.

Similarly, we observe that by the fact that (a_n) is nonincreasing, a₂+ a₃ ≤ 2a₂, a₄+ a₅+ a6+ a7 ≤ 4a₄ and

a₂k+ a₂k+1+ · · · + a₂k+1−1 < 2^ka₂k. Since a_i ≥ 0, we know that

n

X

i=1

a_i= a₁+ · · · + a₂k−1 + · · · + a_n

≤ a₁+ a₂+ a₃+ · · · + a₂k+ a₂k+1+ · · · + a_n+ a_n+1+ · · · + a₂k+1−1

= a1+ (a2+ a3) + · · · + (a₂^k+ a₂^k₊₁+ · · · + a₂^k+1₋₁)

≤ a₁+ 2a2+ · · · + 2^ka₂^k

= a1+

k

X

j=1

2^ja₂^j.

Hence we proved the inequality. 

1

2 JIA-MING (FRANK) LIOU

Corollary 0.1. (Cauchy condensation theorem) Let (an) be a nonnegative non increasing sequence of real numbers. Then either bothP∞

n=1a_nand P∞

n=12ⁿa₂ⁿ 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

n=1

1

n^p is convergent if p > 1 and divergent if p ≤ 1.

Proof. Let an= 1/n^p. Then

2ⁿa₂ⁿ = 2ⁿ· 1

(2ⁿ)^p = 1 2^np−n =

 1 2^p−1

n

. Hence

∞

X

n=1

2ⁿa₂ⁿ =

∞

X

n=1

 1 2^p−1

n

is a geometric series with ratio 1/2^p−1. If p > 1, p − 1 > 0. In this case 2^p−1 > 1 and thus 1/2^p−1 < 1. We know that the infinite seriesP∞

n=1(1/2^p−1)ⁿ is convergent. If p < 1, 1/2^p−1> 1, then the infinite series P∞

n=1(1/2^p−1)ⁿ is divergent. If p = 1, 1/2^p−1 = 1. We know P∞

n=11ⁿ is divergent.

 Similarly, you can determine when the following series is convergent using the Cauchy condensation theorem

∞

X

n=2

1 n(ln n)^p, where ln n = log_en, and e = lim

n→∞

 1 + 1

n

n

.