Advanced book on Mathematics Olympiad

488 Real Analysis

To conclude the solution to the problem, assume that the sequence(an)ndoes not

converge to 0. Then it has some subsequence(ank)kthat approaches a nonzero (finite or

infinite) limit asn→∞. But we saw above that this subsequence has infinitely many

terms in common with a sequence that converges to zero, namely with some(aγk)k.

This is a contradiction. Hence the sequence(an)nconverges to 0.

(Soviet Union University Student Mathematical Olympiad, 1975)

351.The solution follows closely that of the previous problem. Replacingfby|f|we

may assume thatf ≥0. We argue by contradiction. Suppose that there existsa> 0

such that the set

A=f−^1 ((a,∞))={x∈( 0 ,∞)|f(x)>a}

is unbounded. We want to show that there existsx 0 ∈( 0 ,∞)such that the sequence

(nx 0 )n≥ 1 has infinitely many terms inA. The idea is to construct a sequence of closed

intervalsI 1 ⊃I 2 ⊃I 3 ⊃ ···with lengths converging to zero and a sequence of positive

integersn 1 <n 2 <n 3 <···such thatnkIk⊂Afor allk≥1.

LetI 1 be any closed interval inAof length less than 1 and letn 1 =1. Exactly as

in the case of the previous problem, we can show that there exists a positive numberm 1

such that∪m≥m 1 mI 1 is a half-line. Thus there existsn 2 >n 1 such thatn 2 I 1 intersects

A. LetJ 2 be a closed interval of length less than 1 in this intersection. LetI 2 =n^12 J 2.

Clearly,I 2 ⊂I 1 , and the length ofI 2 is less thann^12. Also,n 2 I 2 ⊂A. Inductively, let

nk>nk− 1 be such thatnkIk− 1 intersectsA, and letJkbe a closed interval of length less

than 1 in this intersection. DefineIk=n^1 kJk.

We found the decreasing sequence of intervalsI 1 ⊃I 2 ⊃I 3 ⊃ ···and positive

integersn 1 <n 2 <n 3 <···such thatnkIk⊂A. Cantor’s nested intervals theorem im-

plies the existence of a numberx 0 in the intersection of these intervals. The subsequence

(nkx 0 )klies inA, which means that(nx 0 )nhas infinitely many terms inA. This implies

that the sequencef (nx 0 )does not converge to 0, since it has a subsequence bounded

away from zero. But this contradicts the hypothesis. Hence our assumption was false,

and therefore limx→∞f(x)=0.

Remark.This result is known as Croft’s lemma. It has an elegant proof using the Baire

category theorem.

352.Adding a few terms of the series, we can guess the identity

1

1 +x

+

2

1 +x^2

+···+

2 n

1 +x^2 n

=

1

x− 1

+

2 n+^1

1 −x^2 n+^1

,n≥ 1.

And indeed, assuming that the formula holds forn, we obtain

1

1 +x

+

2

1 +x^2

+···+

2 n

1 +x^2 n

+

2 n+^1

1 +x^2 n+^1

=

1

x− 1

+

2 n+^1

1 −x^2 n+^1

+

2 n+^1

1 +x^2 n+^1